Question

e. \\( 9\\,\text{mol/(l·min)} \\\\ 8. \text{consider the following chemical equations and their respective enthalpy changes. which} \\\\ \ce{n_{2(g)} + 3h_{2(g)} -> 2nh_{3(g)}} \quad \delta h = -91.8\\,\text{kj} \\\\ \ce{c_{(s)} + 2h_{2(g)} -> ch_{4(g)}} \quad \delta h = -74.8\\,\text{kj} \\\\ \ce{h_{2(g)} + 2c_{(s)} + n_{2(g)} -> 2hcn_{(g)}} \quad \delta h = +270.3\\,\text{kj} \\\\ \text{calculate the enthalpy change for the reaction:} \\\\ \ce{ch_{4(g)} + nh_{3(g)} -> hcn_{(g)} + 3h_{2(g)}} \\\\ \text{a. } -437\\,\text{kj} \\\\ \text{b. } +437\\,\text{kj} \\\\ \text{c. } -256\\,\text{kj} \\\\ \text{d. } +256\\,\text{kj} \\\\ \text{e. } \text{none of the above}

Explanation:

Step1: Identify Target Reaction

We need to find \(\Delta H\) for \( \text{CH}_{(g)} + \text{NH}_{3(g)}
ightarrow \text{HCN}_{(g)} + 3\text{H}_{2(g)} \).

Step2: List Given Reactions

1. \( \text{N}_{2(g)} + 3\text{H}_{2(g)}

ightarrow 2\text{NH}_{3(g)} \quad \Delta H = -91.8\ \text{kJ} \) (Reaction 1)

1. \( \text{C}_{(s)} + 2\text{H}_{2(g)}

ightarrow \text{CH}_{4(g)} \quad \Delta H = -74.8\ \text{kJ} \) (Reaction 2)

1. \( \text{H}_{2(g)} + 2\text{C}_{(s)} + \text{N}_{2(g)}

ightarrow 2\text{HCN}_{(g)} \quad \Delta H = +270.3\ \text{kJ} \) (Reaction 3)

Step3: Manipulate Reactions

• Reverse Reaction 1: \( 2\text{NH}_{3(g)}

ightarrow \text{N}_{2(g)} + 3\text{H}_{2(g)} \quad \Delta H = +91.8\ \text{kJ} \) (Reverse, so \(\Delta H\) sign flips)

• Reverse Reaction 2: \( \text{CH}_{4(g)}

ightarrow \text{C}_{(s)} + 2\text{H}_{2(g)} \quad \Delta H = +74.8\ \text{kJ} \) (Reverse, so \(\Delta H\) sign flips)

• Divide Reaction 3 by 2: \( \text{H}_{2(g)} + \text{C}_{(s)} + \frac{1}{2}\text{N}_{2(g)}

ightarrow \text{HCN}_{(g)} \quad \Delta H = +\frac{270.3}{2} = +135.15\ \text{kJ} \) (Reaction 3a)

Step4: Combine Reactions

We need to get \( \text{CH}_{4(g)} + \text{NH}_{3(g)}
ightarrow \text{HCN}_{(g)} + 3\text{H}_{2(g)} \). Let's adjust:

• From reversed Reaction 1 (divided by 2): \( \text{NH}_{3(g)}

ightarrow \frac{1}{2}\text{N}_{2(g)} + \frac{3}{2}\text{H}_{2(g)} \quad \Delta H = +\frac{91.8}{2} = +45.9\ \text{kJ} \) (Reaction 1a)

• Reverse Reaction 2: \( \text{CH}_{4(g)}

ightarrow \text{C}_{(s)} + 2\text{H}_{2(g)} \quad \Delta H = +74.8\ \text{kJ} \) (Reaction 2a)

• Reaction 3a: \( \text{H}_{2(g)} + \text{C}_{(s)} + \frac{1}{2}\text{N}_{2(g)}

ightarrow \text{HCN}_{(g)} \quad \Delta H = +135.15\ \text{kJ} \)

Now, add Reaction 2a, Reaction 1a, and Reaction 3a:

• Reaction 2a: \( \text{CH}_{4(g)}

ightarrow \text{C}_{(s)} + 2\text{H}_{2(g)} \quad \Delta H = +74.8\ \text{kJ} \)

• Reaction 1a: \( \text{NH}_{3(g)}

ightarrow \frac{1}{2}\text{N}_{2(g)} + \frac{3}{2}\text{H}_{2(g)} \quad \Delta H = +45.9\ \text{kJ} \)

• Reaction 3a: \( \text{H}_{2(g)} + \text{C}_{(s)} + \frac{1}{2}\text{N}_{2(g)}

ightarrow \text{HCN}_{(g)} \quad \Delta H = +135.15\ \text{kJ} \)

Step5: Sum Reactions and \(\Delta H\)

• Sum of reactions: \( \text{CH}_{4(g)} + \text{NH}_{3(g)} + \text{H}_{2(g)} + \text{C}_{(s)} + \frac{1}{2}\text{N}_{2(g)}

ightarrow \text{C}_{(s)} + 2\text{H}_{2(g)} + \frac{1}{2}\text{N}_{2(g)} + \frac{3}{2}\text{H}_{2(g)} + \text{HCN}_{(g)} \)

• Simplify: \( \text{CH}_{4(g)} + \text{NH}_{3(g)}

ightarrow \text{HCN}_{(g)} + 3\text{H}_{2(g)} \) (after canceling common terms)

• Sum \(\Delta H\): \( +74.8 + 45.9 + 135.15 = +255.85 \approx +256\ \text{kJ} \)

Answer:

d. +256 kJ