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WorksheetsHess's Law and Bond Enthalpies
Total questions: 15
Worksheet time: 15mins
Based upon the graph above, what type of reaction is this?
Endothermic
Exothermic
Consider the following reactions:Fe(s) + 1/2 O2(g) --> FeO(s) H= -272.0 kJ2 Fe(s) + 3/2 O2(g) --> Fe2O3(s) H= -825.5 kJDetermine the delta H for the reaction: 2 FeO(s) + 1/2 O2(g) --> Fe2O3
-281.5 kJ
-1097.5 kJ
-1369.5 kJ
281.5 kJ
The enthalpy change for the reaction, ∆Hr , is equal to
∆H1 + ∆H2
∆H1 - ∆H2
-∆H1 - ∆H2
-∆H1 + ∆H2
Determine the H for reaction of this equation: 3C (graphite) + 4 H2(g) --> C3H8(g)
using these intermediates:
(A) C3H8 (g) + 5O2(g) --> 3CO2 (g) + 4H2O (l) H= -2219.9 kJ
(B) C(graphite) + O2(g) --> CO2 (g) H= -393.5 kJ
(C) H2 (g) + 1/2 O2 (g) --> H2O (l) H= -285.8 kJ
-683.4 kJ
-2899.2 kJ
-1324.7 kJ
-104.0 kJ
Using the equations below:
C(s) + O2(g) → CO2(g) ∆H = –390 kJ
Mn(s) + O2(g) → MnO2(s) ∆H = –520 kJ
what is ∆H (in kJ) for the following reaction?
MnO2(s) + C(s) → Mn(s) + CO2(g)
910
130
-130
-910
For Hess' Law to be used what must be the same for all of the reactions being studied?
The initial conditions of pressure and temperature.
The final conditions of pressure and temperature.
The initial and the final conditions of pressure and temperature.
The initial and the final conditions of pressure and temperature, and the number of moles of reactants.
Consider the following reactions: \( \ce{H2(g) + 1/2 O2(g) -> H2O(l)} \) \( \Delta H = -285.8 \text{ kJ} \) and \( \ce{C(s) + O2(g) -> CO2(g)} \) \( \Delta H = -393.5 \text{ kJ} \). Calculate \( \Delta H \) for the reaction: \( \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \).
-890.3 kJ
-802.3 kJ
-1000.0 kJ
-750.0 kJ
Which of the following statements best describes Hess's Law?
The total enthalpy change for a reaction is the same, regardless of the number of steps in the reaction.
The enthalpy change of a reaction depends on the path taken.
The enthalpy change of a reaction is always positive.
The enthalpy change of a reaction is always negative.
Given the bond enthalpies: \( \ce{C-H} = 413 \text{ kJ/mol} \), \( \ce{O=O} = 498 \text{ kJ/mol} \), \( \ce{C=O} = 799 \text{ kJ/mol} \), and \( \ce{O-H} = 463 \text{ kJ/mol} \), calculate the \( \Delta H \) for the combustion of methane: \( \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \).
-802 kJ/mol
-890 kJ/mol
-1000 kJ/mol
-750 kJ/mol
Which of the following is a real-world application of Hess's Law?
Determining the energy efficiency of a fuel.
Calculating the speed of a chemical reaction.
Measuring the pH of a solution.
Determining the color change in a reaction.
If the enthalpy change for the reaction \( \ce{N2(g) + 3H2(g) -> 2NH3(g)} \) is \( -92.4 \text{ kJ/mol} \), what is the enthalpy change for the reverse reaction?
92.4 kJ/mol
-92.4 kJ/mol
184.8 kJ/mol
-184.8 kJ/mol
Which of the following reactions would you use Hess's Law to calculate the enthalpy change?
A reaction that occurs in multiple steps.
A reaction that occurs in a single step.
A reaction with no enthalpy change.
A reaction that is at equilibrium.
Using bond enthalpies, calculate the \( \Delta H \) for the reaction: \( \ce{H2(g) + Cl2(g) -> 2HCl(g)} \). Given: \( \ce{H-H} = 436 \text{ kJ/mol} \), \( \ce{Cl-Cl} = 242 \text{ kJ/mol} \), \( \ce{H-Cl} = 431 \text{ kJ/mol} \).
-184 kJ/mol
-92 kJ/mol
-242 kJ/mol
-436 kJ/mol
What is the primary reason for using bond enthalpies in calculations?
To estimate the enthalpy change of a reaction.
To determine the rate of a reaction.
To measure the temperature change in a reaction.
To calculate the pressure change in a reaction.
Which of the following is true about bond enthalpies?
They are average values for a given type of bond.
They are exact values for each specific bond.
They vary significantly with temperature.
They are only applicable to ionic bonds.
