Ultimate Trivia Quiz On Thermochemistry!
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Under the proper conditions, when nitric oxide is allowed to react with chlorine and hydrogen peroxide, hydrogen nitrate and hydrogen chloride are produced as follows: NO(g) + 1/2Cl2(g) + H2O2(g) --> HNO3(g) + HCl(g) Calculate the enthalpy change in kJ for this reaction given the information below: NOCl(g) --> NO(g) + 1/2 Cl2 DH° = 37.0 kJ NOCl(g) + H2O2(g) --> HNO3(g) + HCl(g) DH° = -143.0 kJ
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-106 kJ
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180 kJ
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-180 kJ
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106 kJ
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-90 kJ
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Dive into the Ultimate Trivia Quiz on Thermochemistry! Test your knowledge on reaction enthalpies, entropy changes, and chemical thermodynamics through challenging calculations and conceptual questions. Perfect for students and enthusiasts looking to deepen their understanding of chemical energy changes.
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- 2.
Calculate the enthalpy of reaction for the formation of phosphoric acid from the following unbalanced equation: P4O10 +H2O --> H3PO4 from the following enthalpies of formation: ΔHof(P4O10(s)) = -2984 KJ/mol ΔHof(H2O(l)) = -286 KJ/mol ΔHof(H3PO4(aq)) = -1288 KJ/mol
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-452 kJ/mol
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-4520 kJ/mol
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-8283 kJ/mol
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8283 kj/mol
Correct Answer
A. -452 kJ/molExplanation
The enthalpy of reaction for the formation of phosphoric acid can be calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products. In this case, the enthalpy of reaction is given by:
ΔH = [ΔHof(H3PO4(aq))] - [ΔHof(P4O10(s)) + ΔHof(H2O(l))]
Plugging in the values given:
ΔH = [-1288 KJ/mol] - [-2984 KJ/mol + (-286 KJ/mol)]
Simplifying:
ΔH = -1288 KJ/mol + 2984 KJ/mol + 286 KJ/mol
ΔH = -452 KJ/mol
Therefore, the correct answer is -452 kJ/mol.Rate this question:
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- 3.
Which of the following is a measure of how the order or disorder in a reaction is changing?
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Delta G
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Delta H
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Delta S
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Specific Heat Capacity
Correct Answer
A. Delta SExplanation
Delta S is a measure of how the order or disorder in a reaction is changing. It represents the change in entropy, which is a measure of the randomness or disorder of a system. A positive value of Delta S indicates an increase in disorder, while a negative value indicates a decrease in disorder. Therefore, Delta S is directly related to the change in the order or disorder of a reaction.Rate this question:
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- 4.
The Bond Energy terms for H-H, Cl-Cl, and H-Cl are 440, 240, and 430 kJmol-1 respectively. Calculate the Delta H for the reaction between hydrogen and chlorine gas to produce two moles of hydrogen chloride gas.
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-860
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-620
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-440
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-180
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+240
Correct Answer
A. -180Explanation
The reaction between hydrogen and chlorine gas to produce two moles of hydrogen chloride gas involves breaking one H-H bond (440 kJ/mol) and one Cl-Cl bond (240 kJ/mol), and forming two H-Cl bonds (2 x 430 kJ/mol). The total energy change for the reaction can be calculated by subtracting the energy required to break the bonds from the energy released when the new bonds are formed. In this case, it would be (2 x 430 kJ/mol) - (440 kJ/mol + 240 kJ/mol) = -180 kJ/mol. Therefore, the Delta H for this reaction is -180.Rate this question:
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- 5.
Which of the following equations represents BOTH the standard enthalpy of formation of zinc oxide, AND the standard enthalpy of combustion of zinc?
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Zn(s) + 0.5 O2(g) ==> ZnO(s)
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Zn(g) + 0.5 O2(g) ==> ZnO(s)
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Zn(s) + O(g) ==> ZnO(s)
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Zn(g) + O(g) ==> ZnO(s)
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None of the above
Correct Answer
A. Zn(s) + 0.5 O2(g) ==> ZnO(s)Explanation
The correct answer is Zn(s) + 0.5 O2(g) ==> ZnO(s) because it represents the formation of zinc oxide from its elements (Zn and O2) in their standard states. The standard enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. Additionally, this equation also represents the combustion of zinc because it involves the reaction of zinc with oxygen to form zinc oxide. The standard enthalpy of combustion is the change in enthalpy when one mole of a substance is completely burned in oxygen.Rate this question:
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- 6.
Consider the following reaction profiles. Which statement is true for profile X?
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The reaction is endothermic since the products have a lower enthalpy than the reactants
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The reaction is endothermic since the products have a higher enthalpy than the reactants
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The reaction is exothermic since the products have a lower enthalpy than the reactants
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The reaction is exothermic since the products have a higher enthalpy than the reactants
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Delta H for the reaction is zero
Correct Answer
A. The reaction is endothermic since the products have a higher enthalpy than the reactantsExplanation
In an endothermic reaction, the products have a higher enthalpy than the reactants, meaning that energy is absorbed from the surroundings. This is consistent with profile X, where the energy of the products is higher than the energy of the reactants. Therefore, the correct statement is "The reaction is endothermic since the products have a higher enthalpy than the reactants."Rate this question:
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- 7.
When burning a solid hydrocarbon, which set of values would be most likely?
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Negative Delta H, Negative Delta S, Negative Delta G
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Negative Delta H, Negative Delta S, Positive Delta G
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Negative Delta H, Positive Delta S, Negative Delta G
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Positive Delta H, Negative Delta S, Negative Delta G
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None of the above
Correct Answer
A. Negative Delta H, Positive Delta S, Negative Delta GExplanation
When burning a solid hydrocarbon, the reaction is exothermic (releasing heat), which results in a negative change in enthalpy (Delta H). The combustion process also involves an increase in disorder or randomness (positive change in entropy, Delta S) as the solid hydrocarbon is converted into gaseous products. Finally, since the reaction is exothermic and the increase in disorder is favorable, the change in Gibbs free energy (Delta G) will also be negative. Therefore, the set of values that would be most likely in this scenario is Negative Delta H, Positive Delta S, Negative Delta G.Rate this question:
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- 8.
Given the following information, calculate Delta G (standard) for the reaction below at 25 degrees Celcius. SiCl2(l)+ 2H2O(l) ---> SnO2(s) + 4HCl(g) Delta H = 133.0 KJ Delta S = 401.5 J/K
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-252.6 kJ
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-13.9 kJ
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13.4 kJ
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122.9 kJ
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252.6 kJ
Correct Answer
A. 13.4 kJExplanation
The given answer of 13.4 kJ is incorrect. The correct answer is -252.6 kJ. This can be calculated using the equation ΔG = ΔH - TΔS, where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin. Plugging in the given values, we get ΔG = 133.0 kJ - (25 + 273) K * (401.5 J/K / 1000 J/K) = -252.6 kJ.Rate this question:
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- 9.
Metal has a specific heat capacity of .899 Jg-1K-1 How much energy is required to raise the temperature of 2.00 kg of the metal by 80 K?
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80 kJ
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35.96 kJ
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179800 kJ
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143.84 J
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143.84 kJ
Correct Answer
A. 143.84 kJExplanation
The specific heat capacity of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by 1 Kelvin. In this question, we are given the specific heat capacity of the metal as 0.899 Jg-1K-1. To find the total energy required to raise the temperature of 2.00 kg of the metal by 80 K, we can use the formula: energy = mass x specific heat capacity x change in temperature. Plugging in the values, we get: energy = 2.00 kg x 0.899 Jg-1K-1 x 80 K = 143.84 kJ. Therefore, the correct answer is 143.84 kJ.Rate this question:
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- 10.
All of the following have Delta G standard (of formation) = 0 EXCEPT
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O2(g)
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Br2(g)
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H2(g)
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Ca (s)
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Hg (l)
Correct Answer
A. Br2(g)Explanation
The correct answer is Br2(g) because it has a non-zero standard Gibbs free energy of formation. The standard Gibbs free energy of formation is a measure of the change in free energy when one mole of a substance is formed from its elements in their standard states. For substances with a standard Gibbs free energy of formation equal to zero, it means that they are in their most stable form at standard conditions. O2(g), H2(g), Ca(s), and Hg(l) all have a standard Gibbs free energy of formation equal to zero, indicating that they are in their most stable form. However, Br2(g) does not have a standard Gibbs free energy of formation equal to zero, suggesting that it is not in its most stable form at standard conditions.Rate this question:
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Mar 21, 2023Quiz Edited by
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May 26, 2012Quiz Created by
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