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WorksheetsHess's Law
Total questions: 25
Worksheet time: 25mins
Calculate the ∆H for the following reaction: 2H2O2 → 2H2O + 1 O2 You are given these two equations: 2H2 + O2 → 2H2O ∆H = -572 kJ H2 + O2 → H2O2 ∆H = -188 kJ
∆H = -948 kJ
∆H = -196 kJ
∆H = -384 kJ
∆H = -188 kJ
When you flip a chemical reaction using Hess's Law, what is done to the heat of reaction value?
nothing
the sign is flipped
1/delta H
Using the data below, which is the correct value for the standard enthalpy of formation for TiCl4(l)?
−1538 kJ mol−1
−1094 kJ mol−1
−750 kJ mol−1
+286 kJ mol−1
The enthalpies of combustion of C(s), H2(g) and C4H9OH(l) (in kJmol-1) are as follows
C(s) + O2(g) -> CO2(g) ∆H=a
H2(g) + ½O2(g) -> H2O(l) ∆H=b
C4H9OH(l) + 6O2(g) -> 4CO2(g) + 5H2O(l) ∆H=c
What is the enthalpy change for the reaction shown below?
4C(g) + 5H2(l) + ½O2(g) -> C4H9OH(l)
c – 4a – 5b
2a + 10b - c
4a + 5b - c
2a + 5b + c
When you multiply a chemical equation by 2, what is done to the heat of reaction?
multiplied by 1/2
multiplied by 2
nothing
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: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
The thermochemical equation for the incomplete combustion of carbon into carbon monoxide is represented as: 2C (s) + O2(g) --> 2CO(g) Compute for the H of the reaction if the overall process above can occur in two steps whose thermochemical equations are given below: (1) C (s) + O2(g) --> CO2(g) H= -392.5 kJ (2) 2 CO(g) + O2(g) --> 2CO2(g) H= -566.0 kJ
172.5 kJ
1,353 kJ
-221.0 kJ
959.5 kJ
Determine the H for reaction of this equation: 3C (graphite) + 4 H2(g) --> C3H8(g)
using these intermediates:
C3H8 (g) + 5O2(g) --> 3CO2 (g) + 4H2O (l) H= -2219.9 kJ
C(graphite) + O2(g) --> CO2 (g) H= -393.5 kJ
H2 (g) + 1/2 O2 (g) --> H2O (l) H= -285.8 kJ
-683.4 kJ
-2899.2 kJ
-1324.7 kJ
-104.0 kJ
Calculate the \( \Delta H \) for the following reaction: \( \ce{2H2O2 -> 2H2O + O2} \). You are given these two equations: \( \ce{2H2 + O2 -> 2H2O} \) \( \Delta H = -572 \text{ kJ} \) and \( \ce{H2 + O2 -> H2O2} \) \( \Delta H = -188 \text{ kJ} \)
\( \Delta H = -948 \text{ kJ} \)
\( \Delta H = -196 \text{ kJ} \)
\( \Delta H = -384 \text{ kJ} \)
\( \Delta H = -188 \text{ kJ} \)
When you flip a chemical reaction using Hess's Law, what is done to the heat of reaction value?
nothing
the sign is flipped
1/\( \Delta H \)
The enthalpies of combustion of \( \ce{C(s)} \), \( \ce{H2(g)} \) and \( \ce{C4H9OH(l)} \) (in \( \text{kJ mol}^{-1} \)) are as follows: \( \ce{C(s) + O2(g) -> CO2(g)} \) \( \Delta H = a \), \( \ce{H2(g) + 1/2O2(g) -> H2O(l)} \) \( \Delta H = b \), \( \ce{C4H9OH(l) + 6O2(g) -> 4CO2(g) + 5H2O(l)} \) \( \Delta H = c \). What is the enthalpy change for the reaction shown below? \( \ce{4C(g) + 5H2(l) + 1/2O2(g) -> C4H9OH(l)} \)
\( c - 4a - 5b \)
\( 2a + 10b - c \)
\( 4a + 5b - c \)
\( 2a + 5b + c \)
When you multiply a chemical equation by 2, what is done to the heat of reaction?
multiplied by 1/2
multiplied by 2
nothing
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: \( \ce{Fe(s) + 1/2 O2(g) -> FeO(s)} \) \( \Delta H = -272.0 \text{ kJ} \) and \( \ce{2 Fe(s) + 3/2 O2(g) -> Fe2O3(s)} \) \( \Delta H = -825.5 \text{ kJ} \). Determine the \( \Delta H \) for the reaction: \( \ce{2 FeO(s) + 1/2 O2(g) -> Fe2O3} \)
\( -281.5 \text{ kJ} \)
\( -1097.5 \text{ kJ} \)
\( -1369.5 \text{ kJ} \)
\( 281.5 \text{ kJ} \)
The thermochemical equation for the incomplete combustion of carbon into carbon monoxide is represented as: \( \ce{2C(s) + O2(g) -> 2CO(g)} \). Compute for the \( \Delta H \) of the reaction if the overall process above can occur in two steps whose thermochemical equations are given below: (1) \( \ce{C(s) + O2(g) -> CO2(g)} \) \( \Delta H = -392.5 \text{ kJ} \) (2) \( \ce{2CO(g) + O2(g) -> 2CO2(g)} \) \( \Delta H = -566.0 \text{ kJ} \)
\( 172.5 \text{ kJ} \)
\( 1353 \text{ kJ} \)
\( -221.0 \text{ kJ} \)
\( 959.5 \text{ kJ} \)
Determine the \( \Delta H \) for the reaction of this equation: \( \ce{3C(graphite) + 4H2(g) -> C3H8(g)} \) using these intermediates: \( \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)} \) \( \Delta H = -2219.9 \text{ kJ} \), \( \ce{C(graphite) + O2(g) -> CO2(g)} \) \( \Delta H = -393.5 \text{ kJ} \), \( \ce{H2(g) + 1/2 O2(g) -> H2O(l)} \) \( \Delta H = -285.8 \text{ kJ} \)
\( -683.4 \text{ kJ} \)
\( -2899.2 \text{ kJ} \)
\( -1324.7 \text{ kJ} \)
\( -104.0 \text{ kJ} \)
Which of the following statements best describes Hess's Law?
The total enthalpy change for a reaction is the sum of all changes, regardless of the multiple stages or steps of the reaction.
The enthalpy change of a reaction is independent of the physical states of the reactants and products.
The enthalpy change of a reaction is dependent on the rate of the reaction.
The enthalpy change of a reaction is always positive.
In an experiment, the enthalpy change for the formation of water from hydrogen and oxygen is measured. If the reaction is reversed, what happens to the enthalpy change?
It remains the same.
It becomes zero.
It changes sign.
It doubles.
Which of the following is a real-world application of Hess's Law?
Calculating the energy required to heat a home.
Determining the energy content of foods.
Predicting the weather.
Designing a car engine.
If a reaction is carried out in multiple steps, how does Hess's Law help in determining the overall enthalpy change?
By adding the enthalpy changes of each step.
By multiplying the enthalpy changes of each step.
By averaging the enthalpy changes of each step.
By ignoring the enthalpy changes of each step.
Which of the following reactions would you use Hess's Law to calculate the enthalpy change?
A reaction that occurs in a single step.
A reaction with an unknown enthalpy change that can be broken down into known steps.
A reaction that does not involve any enthalpy change.
A reaction that is already balanced.
What is the significance of Hess's Law in thermochemistry?
It allows the calculation of enthalpy changes for reactions that are difficult to measure directly.
It provides a method to calculate the entropy of a system.
It helps in determining the pressure of a gas.
It is used to calculate the volume of a liquid.
In a laboratory setting, how can Hess's Law be used to determine the enthalpy change of a reaction that cannot be measured directly?
By using a calorimeter to measure the temperature change.
By using known enthalpy changes of related reactions to calculate the unknown enthalpy change.
By measuring the pressure change during the reaction.
By observing the color change of the reactants.
