Question

Calculate the enthalpy change when 6 g. \(\text{CO(g)}\) reacts with sufficient \(\text{NO}_2\text{(g)}\) according to the following reaction
\(4\,\text{CO(g)}+2\,\text{NO}_2\text{(g)}\rightarrow 4\,\text{CO}_2\text{(g)}+2\text{N}_2\text{(g)}\,; \Delta _rH^0 = -1200 \text{kJ}\)

• A. \(-16.08 \text{kJ}\).
• B. \(-32.15 \text{kJ}\).
• C. \(-64.29 \text{kJ}\)
• D. \(-128.58 \text{kJ}\).

Hint:

The given enthalpy is for 4 mol of CO, so scale it to the moles in 6 g.

Answer 1

The Correct Option is C

Step 1: Proportion
Four moles of CO is $4 \times 28 = 112$ g, and this gives $-1200$ kJ.

Step 2: Scale down
For 6 g: $\Delta H = -1200 \times \frac{6}{112} = -\frac{7200}{112} = -64.29$ kJ.

Step 3: Result
This is option (C).

Final Answer:
Enthalpy change for 6 g CO is -64.29 kJ. \[ \boxed{\text{(C)}\ -64.29\ \text{kJ}} \]