Question

1. consider the following chemical equations and their respective enthalpy changes:which

n₂(g) + 3h₂(g) → 2nh₃(g) δh = -91.8 kj
c(s) + 2h₂(g) → ch₄(g). δh = -74.9 kj
h₂(g) + 2c(s) + n₂(g) → 2hcn. δh = 270.3 kj

calculate the enthalpy change for the reaction:
ch₄(g) + nh₃(g) → hcn(g) + 3h₂(g)

a. −437 kj

b. +437 kj

c. −256 kj

d. +256 kj

e. none of the above

Explanation:

Step1: Label the given reactions

Let's call the reactions:

1. \( \ce{N2(g) + 3H2(g) -> 2NH3(g)} \quad \Delta H_1 = -91.8 \, \text{kJ} \)
2. \( \ce{C(s) + 2H2(g) -> CH4(g)} \quad \Delta H_2 = -74.9 \, \text{kJ} \)
3. \( \ce{H2(g) + 2C(s) + N2(g) -> 2HCN(g)} \quad \Delta H_3 = 270.3 \, \text{kJ} \)

We need to find \( \Delta H \) for \( \ce{CH4(g) + NH3(g) -> HCN(g) + 3H2(g)} \)

Step2: Manipulate reaction 1

Reverse reaction 1 and divide by 2:
\( \ce{NH3(g) -> \frac{1}{2}N2(g) + \frac{3}{2}H2(g)} \quad \Delta H_{1}' = -\frac{\Delta H_1}{2} = -\frac{-91.8}{2} = 45.9 \, \text{kJ} \)

Step3: Manipulate reaction 2

Reverse reaction 2:
\( \ce{CH4(g) -> C(s) + 2H2(g)} \quad \Delta H_{2}' = -\Delta H_2 = 74.9 \, \text{kJ} \)

Step4: Manipulate reaction 3

Divide reaction 3 by 2:
\( \ce{\frac{1}{2}H2(g) + C(s) + \frac{1}{2}N2(g) -> HCN(g)} \quad \Delta H_{3}' = \frac{\Delta H_3}{2} = \frac{270.3}{2} = 135.15 \, \text{kJ} \)

Step5: Add the manipulated reactions

Now, add the three manipulated reactions:

• From \( \Delta H_{1}' \): \( \ce{NH3(g) -> \frac{1}{2}N2(g) + \frac{3}{2}H2(g)} \)
• From \( \Delta H_{2}' \): \( \ce{CH4(g) -> C(s) + 2H2(g)} \)
• From \( \Delta H_{3}' \): \( \ce{\frac{1}{2}H2(g) + C(s) + \frac{1}{2}N2(g) -> HCN(g)} \)

Adding them together:
\( \ce{CH4(g) + NH3(g) + \frac{1}{2}H2(g) + C(s) + \frac{1}{2}N2(g) -> \frac{1}{2}N2(g) + \frac{3}{2}H2(g) + C(s) + 2H2(g) + HCN(g)} \)

Simplify the reactants and products (cancel \( C(s) \), \( \frac{1}{2}N2(g) \)):
\( \ce{CH4(g) + NH3(g) -> HCN(g) + 3H2(g)} \) (which is the target reaction)

Step6: Calculate total \( \Delta H \)

Sum the \( \Delta H \) values of the manipulated reactions:
\( \Delta H = \Delta H_{1}' + \Delta H_{2}' + \Delta H_{3}' \)
\( \Delta H = 45.9 + 74.9 + 135.15 \)
\( \Delta H = 255.95 \approx 256 \, \text{kJ} \) (positive because the reaction is endothermic in the direction we want)

Answer:

d. + 256 kJ