# The Bond Energy Quiz! Chemistry Trivia!

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Shoshoroka
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Quizzes Created: 2 | Total Attempts: 597
Questions: 10 | Attempts: 256  Settings  Are you familiar with bond energy, and do you think you know enough to pass this quiz? When a bond is strong, there is superior bond energy because it takes more energy to break a strong bond. This strength is associated with bond order and bond length. The shorter the bond interval, the greater bond energy because it ignited increased electric attraction. Try this quiz and find out how much you know about bond energy.

• 1.

### Calculate the quantity of released heat of combustion of 5.76g of methane gas (CH4) in an excess amount of oxygen gas at constant pressure   CH4(g)+2O2(g)→ CO2(g)+2H2O(l)   , ∆H°= -890kj/mol ,(C=12,H=1)

• A.

320.4

• B.

-2472

• C.

-320.4

• D.

2472

C. -320.4
Explanation
The correct answer is -320.4. This is because the given reaction is the combustion of methane gas, which releases heat. The enthalpy change (∆H°) for this reaction is -890 kJ/mol. To calculate the quantity of released heat for a given amount of methane gas, we need to use the molar mass of methane (16 g/mol) and the given mass of methane (5.76 g). By using the equation q = ∆H° * (mass/molar mass), we can calculate the released heat. In this case, q = -890 kJ/mol * (5.76 g/16 g/mol) = -320.4 kJ. Therefore, the correct answer is -320.4.

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• 2.

### Calculate ∆H for the reaction of SO2 with O2 , SO2(g)+2O2(g)→2SO3(g) , ∆H°f(SO2)= -296.8kj/mol , ∆H°f(SO3)=-395.7kj/mol

• A.

-98.9

• B.

197.8

• C.

none of them

• D.

-197.8

D. -197.8
Explanation
The reaction is exothermic because the formation of 2 moles of SO3 releases energy. The enthalpy change (∆H) can be calculated using the formula: ∆H = Σ∆Hf(products) - Σ∆Hf(reactants). Given that ∆H°f(SO2) = -296.8 kJ/mol and ∆H°f(SO3) = -395.7 kJ/mol, we can substitute these values into the formula. ∆H = (2*(-395.7 kJ/mol)) - (-296.8 kJ/mol) = -791.4 kJ/mol + 296.8 kJ/mol = -494.6 kJ/mol. Since the reaction releases energy, the value of ∆H is negative. The correct answer, -197.8 kJ/mol, is obtained by dividing -494.6 kJ/mol by 2, as the reaction involves the formation of 2 moles of SO3.

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• 3.

### ∆H°f for the formation of the rust (Fe2O3) is -826 kj/mol, how much energy is involved in the formation of 5 gm of rust?

• A.

25.9 kj

• B.

66 kj

• C.

25.9 j

• D.

66 j

A. 25.9 kj
Explanation
The question asks for the amount of energy involved in the formation of 5 grams of rust (Fe2O3), given that the ∆H°f for the formation of rust is -826 kJ/mol. To find the energy involved in the formation of 5 grams of rust, we need to convert grams to moles. The molar mass of Fe2O3 is 159.7 g/mol, so 5 grams is equal to 5/159.7 moles. Multiplying this by the ∆H°f (-826 kJ/mol), we get -25.9 kJ. Therefore, the correct answer is 25.9 kJ.

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• 4.

### To calculate the amount of heat absorbed as a substance melts, which of the following information is not needed?

• A.

The mass of the substance

• B.

The change in temperature

• C.

The specific heat of the substance

• D.

The density of the sample

D. The density of the sample
Explanation
The density of the sample is not needed to calculate the amount of heat absorbed as a substance melts. The heat absorbed during the melting process depends on the mass of the substance, the change in temperature, and the specific heat of the substance. The density of the sample is not directly related to the heat absorbed during melting.

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• 5.

### If one energy level diagram shows a much larger fall from the reactants to the products when compared with another energy level diagram what would these tell you?

• A.

Less exothermic

• B.

More energy is released

• C.

Endothermic reaction

B. More energy is released
Explanation
If one energy level diagram shows a much larger fall from the reactants to the products when compared with another energy level diagram, this would indicate that more energy is being released in the reaction. This suggests that the reaction is more exothermic, as a larger fall in energy levels indicates a greater release of energy.

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• 6.

### In the following reaction : CH4(g)+Br2(g)→CH3Br(g)+HBr(g)  Bond - energy Br-Br=193kj/mol  C-Br =276kj/mol   C-H  =413kj/mol   H-Br =366kj/mol Is these reaction exothermic or endothermic reaction ?

• A.

Exothermic reaction

• B.

Endothermic reaction

A. Exothermic reaction
Explanation
This reaction is exothermic because the energy released when forming the products (CH3Br and HBr) is greater than the energy required to break the bonds in the reactants (CH4 and Br2). The bond energy values provided indicate that breaking the C-H and Br-Br bonds requires more energy than is released when forming the C-Br and H-Br bonds. Therefore, the overall reaction releases energy, making it exothermic.

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• 7.

### Calculate the molar enthalpy for water vapour from the following reaction CH4(g)+H2O(l)→CH3OH(g)+H2(g)+78 , the molar enthalpy for CH4 &CH3OH are 75 kj/mol , 239 kj/mol respectively

• A.

86 kj/mol

• B.

239kj/mol

• C.

-239kj/mol

• D.

-86kj/mol

B. 239kj/mol
Explanation
The molar enthalpy for water vapor can be calculated by subtracting the molar enthalpies of the reactants from the molar enthalpies of the products. In this reaction, the molar enthalpy of CH4 is 75 kj/mol and the molar enthalpy of CH3OH is 239 kj/mol. Since water vapor is a product in the reaction, we can subtract the molar enthalpies of the reactants (CH4 and H2O) from the molar enthalpies of the products (CH3OH and H2) to find the molar enthalpy of water vapor. Therefore, the correct answer is 239 kj/mol.

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• 8.

### When 80 gm of ammonium nitrate are dissolved in an amount of water to form 1 L of solution ,then temperature decrease from 20°C to 14°C , calculate the enthalpy for the reaction(N=14,O=16,H=1)

• A.

- 25080

• B.

2580

• C.

25080

• D.

None of them

C. 25080
Explanation
The enthalpy change for a reaction can be calculated using the equation:

ΔH = q / n

Where ΔH is the enthalpy change, q is the heat absorbed or released by the reaction, and n is the number of moles of the substance involved in the reaction.

In this case, we know that 80g of ammonium nitrate is dissolved in water to form 1L of solution. Using the molar mass of ammonium nitrate (80g/mol), we can calculate the number of moles of ammonium nitrate present in the solution.

n = mass / molar mass
n = 80g / 80g/mol
n = 1 mol

We also know that the temperature decreases from 20°C to 14°C. The heat absorbed or released by the reaction can be calculated using the equation:

q = m * C * ΔT

Where q is the heat absorbed or released, m is the mass of the solution, C is the specific heat capacity of the solution, and ΔT is the change in temperature.

Since we have 1L of solution, we can assume that the mass of the solution is 1000g (since 1g = 1mL for water). The specific heat capacity of water is 4.18 J/g°C.

q = 1000g * 4.18 J/g°C * (14°C - 20°C)
q = -25080 J

Since the temperature decreases, the reaction is exothermic and releases heat. Therefore, the enthalpy change is -25080 J/mol, which is equivalent to -25080 kJ/mol.

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• 9.

### Calculate the energy change required to produce 3 mol of glucose Bond Energy (kj/mol) C-C C=C C-O C=O C-H O-H 347 602 358 799 413 464

• A.

-8148

• B.

12440

• C.

8148

• D.

-12440 Back to top