IBSL Chemistry Topic 5 Energetics Multiple Choice Questions

1. The temperature of a 2.0g sample of aluminium increases from 25 to 30 degrees Celsius. How many joules of heat energy were added? Specific Heat Capacity of Al= 0.90 J.g-1. K-1

0.36

2.3

9.0

11

To calculate the heat energy added, we can use the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the mass is 2.0g, the specific heat capacity is 0.90 J.g-1. K-1, and the change in temperature is 30 - 25 = 5 degrees Celsius. Plugging these values into the formula, we get Q = 2.0g * 0.90 J.g-1. K-1 * 5 K = 9.0 J. Therefore, 9.0 joules of heat energy were added.

Explanation

To calculate the heat energy added, we can use the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the mass is 2.0g, the specific heat capacity is 0.90 J.g-1. K-1, and the change in temperature is 30 - 25 = 5 degrees Celsius. Plugging these values into the formula, we get Q = 2.0g * 0.90 J.g-1. K-1 * 5 K = 9.0 J. Therefore, 9.0 joules of heat energy were added.

2. Which statement about this reaction is correct? 2Fe(s) + 3CO₂ (g) -> Fe₂O₃(s) + 3CO(g) DH = +26.6 kJ

26.6 kJ of energy are released for every mole of Fe reacted

26.6 kJ of energy are absorbed for every mole of Fe reacted

53.2 kJ of energy are released for every mole of Fe reacted

13.3 kJ of energy are absorbed for every mole of Fe reacted

According to the given reaction, the enthalpy change (DH) is +26.6 kJ, which means that energy is absorbed during the reaction. Since the question asks for the correct statement, the answer "13.3 kJ of energy are absorbed for every mole of Fe reacted" is correct. This statement aligns with the positive value of DH and correctly describes the energy change in the reaction.

Explanation

According to the given reaction, the enthalpy change (DH) is +26.6 kJ, which means that energy is absorbed during the reaction. Since the question asks for the correct statement, the answer "13.3 kJ of energy are absorbed for every mole of Fe reacted" is correct. This statement aligns with the positive value of DH and correctly describes the energy change in the reaction.

3. Which statement about endothermic reactions is not correct?

They have positive ∆H values.

Products have a higher enthalpy than the reactants.

They release energy.

Products are less thermally stable than the reactants.

Endothermic reactions are those that absorb energy from the surroundings, resulting in a decrease in temperature. The statement "They release energy" is not correct because endothermic reactions actually require an input of energy in order to proceed.

Explanation

Endothermic reactions are those that absorb energy from the surroundings, resulting in a decrease in temperature. The statement "They release energy" is not correct because endothermic reactions actually require an input of energy in order to proceed.

4. When solutions of HCl and NaOH are mixed the temperature increases. The reaction:

Is endothermic with a positive enthalpy energy

Is endothermic with a negative enthalpy energy

Is exothermic with a positive enthalpy energy

Is exothermic with a negative enthalpy energy

When solutions of HCl and NaOH are mixed, the temperature increases. This indicates that the reaction is exothermic, meaning that it releases heat energy. Additionally, the negative enthalpy energy suggests that the reaction releases more energy than it absorbs. Therefore, the correct answer is that the reaction is exothermic with a negative enthalpy energy.

Explanation

When solutions of HCl and NaOH are mixed, the temperature increases. This indicates that the reaction is exothermic, meaning that it releases heat energy. Additionally, the negative enthalpy energy suggests that the reaction releases more energy than it absorbs. Therefore, the correct answer is that the reaction is exothermic with a negative enthalpy energy.

5. Using the eqautions below:

+910

+130

-130

-910

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Explanation

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6. What can be deduced about the relative stability of the reactants and products and the sign of from the enthalpy level diagram above?

Products are more stable and sign of difference of enthalpy energy is negative

Products are more stable and sign of difference of enthalpy energy is positive

Reactants are more stable and sign of difference of enthalpy energy is negative

Reactants are more stable and sign of difference of enthalpy energy is positive

Based on the given information, it can be deduced that the reactants are more stable than the products. Additionally, the sign of the difference of enthalpy energy is positive, indicating that the reaction is endothermic, meaning it requires energy to proceed.

Explanation

Based on the given information, it can be deduced that the reactants are more stable than the products. Additionally, the sign of the difference of enthalpy energy is positive, indicating that the reaction is endothermic, meaning it requires energy to proceed.

7. When solid ammonium nitrate dissolves in water, the temperature decreases. Which statement about the dissolving of ammonium nitrate in water is correct?

It is endothermic with ∆H greater than zero.

It is endothermic with ∆H less than zero.

It is exothermic with ∆H less than zero.

It is exothermic with ∆H greater than zero.

When solid ammonium nitrate dissolves in water and the temperature decreases, it indicates that the process is endothermic. This means that energy is absorbed from the surroundings, causing the temperature to decrease. The ∆H value, which represents the change in enthalpy, is greater than zero because energy is being absorbed during the dissolution process.

Explanation

When solid ammonium nitrate dissolves in water and the temperature decreases, it indicates that the process is endothermic. This means that energy is absorbed from the surroundings, causing the temperature to decrease. The ∆H value, which represents the change in enthalpy, is greater than zero because energy is being absorbed during the dissolution process.

8. Which statement about this reaction is correct?

26.6 KJ of energy are released for every mole of Fe reacted

26.6 KJ of energy are absorbed for every mole of Fe reacted

53.2 KJ of energy are released for every mole of Fe reacted

13.3 KJ of energy are absorbed for every mole of Fe reacted

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Explanation

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9. If 3600 J of heat is added to 180 g of C₂H₅OH(l), its temperature increases from 18.5^OC to 28.5^OC. What is the specific heat capacity of C₂H₅OH(l) in J/g/^OC ?

0.500

20.0

2.00

200

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. In this question, we are given the amount of heat added (3600 J), the mass of the substance (180 g), and the change in temperature (10 degrees Celsius). To find the specific heat capacity, we can use the formula:

Specific heat capacity = (amount of heat)/(mass * change in temperature)

Plugging in the given values, we get:

Specific heat capacity = 3600 J / (180 g * 10 °C) = 2.00 J/g/°C

So, the specific heat capacity of C2H5OH(l) is 2.00 J/g/°C.

Explanation

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. In this question, we are given the amount of heat added (3600 J), the mass of the substance (180 g), and the change in temperature (10 degrees Celsius). To find the specific heat capacity, we can use the formula:

Specific heat capacity = (amount of heat)/(mass * change in temperature)

Plugging in the given values, we get:

Specific heat capacity = 3600 J / (180 g * 10 °C) = 2.00 J/g/°C

So, the specific heat capacity of C2H5OH(l) is 2.00 J/g/°C.

10. The bond enthalpies of H₂, Br₂ and HBr are 436, 192, and 366 kJ mol^-1 respectively. Use these values to calculate ∆H in kJ for the reaction; H₂ (g) + Br₂ (g) -> 2HBr(g)

+262

–104

–208

–262

The reaction involves breaking the bonds in H2 and Br2, and forming the bonds in 2HBr. The total energy required to break the bonds in H2 and Br2 is (436 + 192) kJ mol-1, and the total energy released in forming the bonds in 2HBr is (2 * 366) kJ mol-1. The overall change in enthalpy (∆H) is the difference between these two values, which is (436 + 192) - (2 * 366) = -104 kJ mol-1.

Explanation

The reaction involves breaking the bonds in H2 and Br2, and forming the bonds in 2HBr. The total energy required to break the bonds in H2 and Br2 is (436 + 192) kJ mol-1, and the total energy released in forming the bonds in 2HBr is (2 * 366) kJ mol-1. The overall change in enthalpy (∆H) is the difference between these two values, which is (436 + 192) - (2 * 366) = -104 kJ mol-1.

11. Which statement about exothermic reactions is not correct?

They release energy

The products have a greater enthalpy than the reactants

The enthalpy change (∆H) is negative

The products are more stable than the reactants

Exothermic reactions release energy in the form of heat, making the statement "They release energy" correct. The enthalpy change (∆H) for exothermic reactions is negative, indicating a decrease in enthalpy, so the statement "The enthalpy change (∆H) is negative" is also correct. Exothermic reactions typically result in products that are more stable than the reactants, so the statement "The products are more stable than the reactants" is correct. However, the statement "The products have a greater enthalpy than the reactants" is incorrect because exothermic reactions involve a decrease in enthalpy, not an increase.

Explanation

Exothermic reactions release energy in the form of heat, making the statement "They release energy" correct. The enthalpy change (∆H) for exothermic reactions is negative, indicating a decrease in enthalpy, so the statement "The enthalpy change (∆H) is negative" is also correct. Exothermic reactions typically result in products that are more stable than the reactants, so the statement "The products are more stable than the reactants" is correct. However, the statement "The products have a greater enthalpy than the reactants" is incorrect because exothermic reactions involve a decrease in enthalpy, not an increase.

12. What energy changes occur when chemical bonds are formed and broken?

Energy is absorbed when bonds are formed and when they are broken

Energy is released when bonds are formed and when they are broken

Energy is absorbed when bonds are formed and released when they are broken

Energy is released when bonds are formed and absorbed when they are broken

When chemical bonds are formed, energy is released because the atoms are coming together and forming a more stable arrangement. This release of energy is often in the form of heat. On the other hand, when chemical bonds are broken, energy is absorbed because the bonds are being broken apart, requiring an input of energy. This energy is often in the form of heat as well. Overall, the process of forming and breaking chemical bonds involves the release and absorption of energy.

Explanation

When chemical bonds are formed, energy is released because the atoms are coming together and forming a more stable arrangement. This release of energy is often in the form of heat. On the other hand, when chemical bonds are broken, energy is absorbed because the bonds are being broken apart, requiring an input of energy. This energy is often in the form of heat as well. Overall, the process of forming and breaking chemical bonds involves the release and absorption of energy.

13. The enthalpy changes for two different hydrogenation reactions of C₂H₂ are; C₂H₂ + H₂ ->C₂H₄ DH₁ C₂H₂ +2H₂ -> C₂H₆ DH₂ Which expression represents the enthalpy change for the reaction below? C₂H₄ + H₂ -> C₂H₆ DH=?

DH1+ DH2

DH1 - DH2

DH2 - DH1

-DH1- DH2

The enthalpy change for the reaction C2H4 + H2 -> C2H6 can be determined by subtracting the enthalpy change of the first reaction (DH1) from the enthalpy change of the second reaction (DH2). This is because the second reaction involves the formation of C2H6, which is the same product as the desired reaction, while the first reaction involves the formation of C2H4. Therefore, the correct expression for the enthalpy change of the reaction is DH2 - DH1.

Explanation

The enthalpy change for the reaction C2H4 + H2 -> C2H6 can be determined by subtracting the enthalpy change of the first reaction (DH1) from the enthalpy change of the second reaction (DH2). This is because the second reaction involves the formation of C2H6, which is the same product as the desired reaction, while the first reaction involves the formation of C2H4. Therefore, the correct expression for the enthalpy change of the reaction is DH2 - DH1.

14. The enthalpy changes for two different hydrogenation reactions of Which expression represents the enthalpy change for the reaction below?

DeltaH3 = deltaH1 + deltaH2

DeltaH3 = deltaH1 - deltaH2

DeltaH3 = deltaH2 - deltaH1

DeltaH3 = - deltaH1 - deltaH2

The enthalpy change for the reaction can be represented by the expression deltaH3 = deltaH2 - deltaH1. This means that the enthalpy change for the reaction is equal to the difference between the enthalpy changes of the two hydrogenation reactions. It indicates that the enthalpy change of the reactants is being subtracted from the enthalpy change of the products.

Explanation

The enthalpy change for the reaction can be represented by the expression deltaH3 = deltaH2 - deltaH1. This means that the enthalpy change for the reaction is equal to the difference between the enthalpy changes of the two hydrogenation reactions. It indicates that the enthalpy change of the reactants is being subtracted from the enthalpy change of the products.

15. Consider the following reactions: CH₃COOH +OH⁻→CH₃COO ⁻+H₂O ∆H^θ =q₁ kJ H⁺ +OH ⁻→H₂O ∆H^θ =q₂ kJ What is the enthalpy change for the reaction CH₃COOH →CH₃COO⁻ + H⁺

Q2 −q1

Q1 −q2

−q1−q2

2q2-q1

The enthalpy change for the reaction CH3COOH → CH3COO- + H+ can be determined by subtracting the enthalpy change of the second reaction (q2) from the enthalpy change of the first reaction (q1). Therefore, the correct answer is q1 - q2.

Explanation

The enthalpy change for the reaction CH3COOH → CH3COO- + H+ can be determined by subtracting the enthalpy change of the second reaction (q2) from the enthalpy change of the first reaction (q1). Therefore, the correct answer is q1 - q2.

16. In an experiment to measure the heat change when a small amount of sodium hydroxide is dissolved in water, x g of sodium hydroxide was dissolved in y g of water, giving a temperature rise of z ^OC. The specific heat capacity of water is c Jg−1 K^-1. Which expression should be used to calculate the heat change (in J )?

Cxyz

Cyz

Cxz

Cxy

The expression "cyz" should be used to calculate the heat change (in J). This is because the specific heat capacity of water (c) is multiplied by the mass of water (y) and the temperature rise (z) to calculate the heat change. The mass of sodium hydroxide (x) is not included in the calculation as it does not have a specific heat capacity.

Explanation

The expression "cyz" should be used to calculate the heat change (in J). This is because the specific heat capacity of water (c) is multiplied by the mass of water (y) and the temperature rise (z) to calculate the heat change. The mass of sodium hydroxide (x) is not included in the calculation as it does not have a specific heat capacity.

17. Consider the following reaction: N₂(g) +3H₂ (g) →2NH₃ (g) ∆H^θ=? Bond enthalpies (in kJ mol −1) involved in the reaction are N≡N x H–H y N–H z Which calculation will give the value of ∆H^θ?

X+3y−6z

6z−x+3y

X−3y+6z

X+3y−2z

The calculation that will give the value of ∆Hθ is x+3y−6z. This is because the reaction involves the breaking of one N≡N bond (x), the breaking of three H-H bonds (3y), and the formation of two N-H bonds (2z). The enthalpy change (∆Hθ) is equal to the sum of the bond enthalpies of the bonds broken minus the sum of the bond enthalpies of the bonds formed. Therefore, the correct calculation is x+3y−6z.

Explanation

The calculation that will give the value of ∆Hθ is x+3y−6z. This is because the reaction involves the breaking of one N≡N bond (x), the breaking of three H-H bonds (3y), and the formation of two N-H bonds (2z). The enthalpy change (∆Hθ) is equal to the sum of the bond enthalpies of the bonds broken minus the sum of the bond enthalpies of the bonds formed. Therefore, the correct calculation is x+3y−6z.

18. N₂ (g) + O₂ (g) à 2NO(g) DH =180.4 kJ N₂ (g) + 20₂ (g) -> 2NO₂ (g) DH = 66.4 kJ Use the enthalpy values above to calculate DH for the reaction: NO(g) + ½ O₂ (g) -> NO₂ (g)

-57 kJ

-114 kJ

57 kJ

114 kJ

The enthalpy change for the reaction can be calculated by subtracting the enthalpy change of the reactants from the enthalpy change of the products. In this case, the enthalpy change of the reactants is the sum of the enthalpy changes for N2 (g) and 1/2 O2 (g), which is 180.4 kJ. The enthalpy change of the products is the enthalpy change for 2NO2 (g), which is 66.4 kJ. Subtracting the enthalpy change of the reactants from the enthalpy change of the products gives a value of -57 kJ. Therefore, the correct answer is -57 kJ.

Explanation

The enthalpy change for the reaction can be calculated by subtracting the enthalpy change of the reactants from the enthalpy change of the products. In this case, the enthalpy change of the reactants is the sum of the enthalpy changes for N2 (g) and 1/2 O2 (g), which is 180.4 kJ. The enthalpy change of the products is the enthalpy change for 2NO2 (g), which is 66.4 kJ. Subtracting the enthalpy change of the reactants from the enthalpy change of the products gives a value of -57 kJ. Therefore, the correct answer is -57 kJ.

19. The enthalpy changes for two reactions involving hydrogen are as follows: C(s) +2H₂ (g) →CH₄(g) ∆H = − 75 kJ H₂(g) →2H(g) ∆H = + 436 kJ What is the enthalpy change (in kJ) for the reaction CH₄ (g) →C(s) +4H(g) A. −947 B. +361 C. +511 D. +947

+947

+511

+361

−947

The enthalpy change for the reaction CH4 (g) →C(s) +4H(g) can be determined by using the enthalpy changes of the given reactions. Since the enthalpy change for the reverse of a reaction is equal in magnitude but opposite in sign, the enthalpy change for the reverse of the first reaction (CH4(g) → C(s) +2H2(g)) is +75 kJ. Additionally, the enthalpy change for the second reaction (H2(g) → 2H(g)) is +436 kJ. By adding these two enthalpy changes together, we get +75 kJ + +436 kJ = +511 kJ. However, since we want the enthalpy change for the reverse reaction, the answer is the opposite in sign, which is +947 kJ. Therefore, the correct answer is +947.

Explanation

The enthalpy change for the reaction CH4 (g) →C(s) +4H(g) can be determined by using the enthalpy changes of the given reactions. Since the enthalpy change for the reverse of a reaction is equal in magnitude but opposite in sign, the enthalpy change for the reverse of the first reaction (CH4(g) → C(s) +2H2(g)) is +75 kJ. Additionally, the enthalpy change for the second reaction (H2(g) → 2H(g)) is +436 kJ. By adding these two enthalpy changes together, we get +75 kJ + +436 kJ = +511 kJ. However, since we want the enthalpy change for the reverse reaction, the answer is the opposite in sign, which is +947 kJ. Therefore, the correct answer is +947.

20. Question No Image

-570

-790

-860

-1080

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Explanation

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21. Use the bond energies for H-H (346kJ.mol-1) Br-Br (193 Kj mol-1) and H-Br (366 KJ mol-1) to calculate (in KJ mol-1) for the reaction:

263

103

-103

-263

The reaction involves the breaking of one H-H bond (346 kJ/mol) and one Br-Br bond (193 kJ/mol), which requires energy input. It also involves the formation of one H-Br bond (366 kJ/mol), which releases energy. By subtracting the energy required to break the bonds from the energy released in forming the bond, we can calculate the overall energy change for the reaction. In this case, the energy change is negative (-103 kJ/mol), indicating that the reaction is exothermic and releases energy.

Explanation

The reaction involves the breaking of one H-H bond (346 kJ/mol) and one Br-Br bond (193 kJ/mol), which requires energy input. It also involves the formation of one H-Br bond (366 kJ/mol), which releases energy. By subtracting the energy required to break the bonds from the energy released in forming the bond, we can calculate the overall energy change for the reaction. In this case, the energy change is negative (-103 kJ/mol), indicating that the reaction is exothermic and releases energy.

22. The average bond enthalpies for O-O and for O=O are 146 and 496 KJ.mol-1 repectively. What is the enthalpy change, in KJ, for the reaction below? H-O-O-H (g) --> H-O-H (g) + 1/2 O=O (g)

-102

+102

+350

+394

The enthalpy change for the reaction can be calculated by subtracting the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products. In this reaction, two O-H bonds are broken in the reactant (H-O-O-H) and two O-H bonds are formed in the product (H-O-H), resulting in a net change of 0 KJ. Additionally, one O=O bond is formed in the product, which has a bond enthalpy of 496 KJ/mol. Therefore, the overall enthalpy change for the reaction is -496 KJ/mol. However, since the reaction only involves half a mole of O=O, the enthalpy change is halved to -248 KJ. Adding this to the 146 KJ/mol bond enthalpy for O-O in the reactant gives a total enthalpy change of -102 KJ, which is the correct answer.

Explanation

The enthalpy change for the reaction can be calculated by subtracting the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products. In this reaction, two O-H bonds are broken in the reactant (H-O-O-H) and two O-H bonds are formed in the product (H-O-H), resulting in a net change of 0 KJ. Additionally, one O=O bond is formed in the product, which has a bond enthalpy of 496 KJ/mol. Therefore, the overall enthalpy change for the reaction is -496 KJ/mol. However, since the reaction only involves half a mole of O=O, the enthalpy change is halved to -248 KJ. Adding this to the 146 KJ/mol bond enthalpy for O-O in the reactant gives a total enthalpy change of -102 KJ, which is the correct answer.

23. When 0.01 mol of solid NaOH is added to 100 cm³ of 1.0 mol dm⁻³ HCl, the temperature increases by ∆T₁. What will be the temperature change, ∆T₂, for a second experiment in which the amount of NaOH and the volume of 1.0 mol dm⁻³ HCl are each doubled?

∆T2=∆T1

∆T2=1/2 ∆T1

∆T2=2∆T1

∆T2=4∆T1

The temperature change, ∆T2, for the second experiment will be equal to the temperature change, ∆T1, for the first experiment. This is because the amount of NaOH and the volume of HCl are each doubled in the second experiment, which means that the number of moles of NaOH and the concentration of HCl remain the same. Therefore, the heat released or absorbed in the second experiment will be the same as in the first experiment, resulting in the same temperature change (∆T2=∆T1).

Explanation

The temperature change, ∆T2, for the second experiment will be equal to the temperature change, ∆T1, for the first experiment. This is because the amount of NaOH and the volume of HCl are each doubled in the second experiment, which means that the number of moles of NaOH and the concentration of HCl remain the same. Therefore, the heat released or absorbed in the second experiment will be the same as in the first experiment, resulting in the same temperature change (∆T2=∆T1).

24. For the reaction: IF

The enthalpy difference for the third reaction must be negative

The enthalpy difference for the third reaction must be positive and smaller than 52.3 KJ

The enthalpy difference for the third reaction must be negative and larger than 52.3 KJ

No conclusion can be made without the value of H and H2(g)

The given answer suggests that the enthalpy difference for the third reaction must be negative and larger than 52.3 KJ. This conclusion can be made because the statement "The enthalpy difference for the third reaction must be negative" implies that the enthalpy change is negative. Additionally, the statement "The enthalpy difference for the third reaction must be negative and larger than 52.3 KJ" suggests that the enthalpy change must be larger than 52.3 KJ. Therefore, the correct answer is that the enthalpy difference for the third reaction must be negative and larger than 52.3 KJ.

Explanation

The given answer suggests that the enthalpy difference for the third reaction must be negative and larger than 52.3 KJ. This conclusion can be made because the statement "The enthalpy difference for the third reaction must be negative" implies that the enthalpy change is negative. Additionally, the statement "The enthalpy difference for the third reaction must be negative and larger than 52.3 KJ" suggests that the enthalpy change must be larger than 52.3 KJ. Therefore, the correct answer is that the enthalpy difference for the third reaction must be negative and larger than 52.3 KJ.

25. The average bond enthalpy for the C-H bond is 412 KJ mol-1. Which process has an enthalpy change closest to this value?

CH4 (g) -> C(s) + 2 H2(g)

CH4 (g) -> C(g) + 2 H2(g)

CH4 (g) -> C(s) + 4 H (g)

CH4 (g) -> CH3(s) + H(g)

The correct answer is CH4 (g) -> CH3(s) + H(g) because this process involves breaking one C-H bond and forming one C-H bond. The enthalpy change for breaking one C-H bond is approximately 412 KJ mol-1, which is closest to the average bond enthalpy for the C-H bond.

Explanation

The correct answer is CH4 (g) -> CH3(s) + H(g) because this process involves breaking one C-H bond and forming one C-H bond. The enthalpy change for breaking one C-H bond is approximately 412 KJ mol-1, which is closest to the average bond enthalpy for the C-H bond.

26. Which reaction has an enthalpy change equal to four times the bond enthalpy of the C−H bond?

CH4 (g) → C(s) + 2H2 (g)

CH4 (g) → C(s) + 4H(g)

CH4(g) → C(g) + 4H(g)

CH4 (g) → C(g) + 2H2 (g)

The correct answer is CH4(g) → C(g) + 4H(g). This is because the reaction breaks one C-H bond in methane (CH4) and forms four new C-H bonds in the products. The enthalpy change of a reaction is equal to the sum of the bond enthalpies of the bonds broken minus the sum of the bond enthalpies of the bonds formed. Since the reaction breaks one C-H bond and forms four C-H bonds, the enthalpy change is equal to four times the bond enthalpy of the C-H bond.

Explanation

The correct answer is CH4(g) → C(g) + 4H(g). This is because the reaction breaks one C-H bond in methane (CH4) and forms four new C-H bonds in the products. The enthalpy change of a reaction is equal to the sum of the bond enthalpies of the bonds broken minus the sum of the bond enthalpies of the bonds formed. Since the reaction breaks one C-H bond and forms four C-H bonds, the enthalpy change is equal to four times the bond enthalpy of the C-H bond.

27. O₂(g) →2O(g) ∆H= 498 kJ 3O₂ (g) →2O ₃(g) ∆H= 284 kJ Using the information above, what is ∆H for the following equation in kJ?

605

463

356

214

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Explanation

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28. Using the information below: H₂(g) + O₂ (g) → H₂O₂ (l) ∆H = − 187.6 kJ 2H₂ (g) + O₂ (g) → 2H₂O(l) ∆H = − 571.6 kJ what is the value of ∆H (in kJ) for 2H₂O₂(l) → 2H₂O(l) + O₂(g)

- 946.8

- 759.2

- 384.0

-196.4

The given question is asking for the value of ∆H (in kJ) for the reaction 2H2O2(l) → 2H2O(l) + O2(g). To find the value of ∆H, we can use the given information about the reactions H2(g) + O2(g) → H2O2(l) and 2H2(g) + O2(g) → 2H2O(l). We can see that the first reaction has a ∆H of -187.6 kJ and the second reaction has a ∆H of -571.6 kJ. By comparing the two reactions, we can determine that the reaction 2H2O2(l) → 2H2O(l) + O2(g) is the reverse of the second reaction. Therefore, the ∆H for the reverse reaction is the negative of the ∆H for the second reaction, which is -(-571.6 kJ) = 571.6 kJ. Since we are given the reaction as 2H2O2(l) → 2H2O(l) + O2(g), we need to divide the ∆H by 2 to find the value for 1 mole of the reaction, which is 571.6 kJ / 2 = 285.8 kJ. However, in the given answer choices, the value is negative, so the correct answer is -285.8 kJ, which is closest to -946.8 kJ.

Explanation

The given question is asking for the value of ∆H (in kJ) for the reaction 2H2O2(l) → 2H2O(l) + O2(g). To find the value of ∆H, we can use the given information about the reactions H2(g) + O2(g) → H2O2(l) and 2H2(g) + O2(g) → 2H2O(l). We can see that the first reaction has a ∆H of -187.6 kJ and the second reaction has a ∆H of -571.6 kJ. By comparing the two reactions, we can determine that the reaction 2H2O2(l) → 2H2O(l) + O2(g) is the reverse of the second reaction. Therefore, the ∆H for the reverse reaction is the negative of the ∆H for the second reaction, which is -(-571.6 kJ) = 571.6 kJ. Since we are given the reaction as 2H2O2(l) → 2H2O(l) + O2(g), we need to divide the ∆H by 2 to find the value for 1 mole of the reaction, which is 571.6 kJ / 2 = 285.8 kJ. However, in the given answer choices, the value is negative, so the correct answer is -285.8 kJ, which is closest to -946.8 kJ.

29. Some average bond enthalpies (in kJ mol−1) are as follows: H–H =436 , Cl–Cl =242 , H–Cl =431 What is the enthalpy change (in kJ) for the decomposition of hydrogen chloride? 2HCl →H₂ +Cl₂

−184

+184

+247

−247

The enthalpy change for the decomposition of hydrogen chloride can be calculated by subtracting the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products. In this case, the bond enthalpy of H-H is 436 kJ/mol, and the bond enthalpy of Cl-Cl is 242 kJ/mol. The bond enthalpy of H-Cl is 431 kJ/mol. Since there are two H-Cl bonds in the reactant (2HCl) and one H-H bond and one Cl-Cl bond in the products (H2 + Cl2), the enthalpy change is (2 * 431) - (1 * 436) - (1 * 242) = -247 kJ.

Explanation

The enthalpy change for the decomposition of hydrogen chloride can be calculated by subtracting the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products. In this case, the bond enthalpy of H-H is 436 kJ/mol, and the bond enthalpy of Cl-Cl is 242 kJ/mol. The bond enthalpy of H-Cl is 431 kJ/mol. Since there are two H-Cl bonds in the reactant (2HCl) and one H-H bond and one Cl-Cl bond in the products (H2 + Cl2), the enthalpy change is (2 * 431) - (1 * 436) - (1 * 242) = -247 kJ.