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: Read the reaction
The equation shows \(4\) mol of \(CO(g)\) reacting, with \(\Delta_rH^0 = -1200\ \text{kJ}\). So the enthalpy change per mole of CO is \(\frac{-1200}{4} = -300\ \text{kJ}\).

Step 2: Moles of CO in 6 g
Molar mass of CO is \(28\ \text{g/mol}\).
\[ n = \frac{6}{28} = 0.2143\ \text{mol} \]

Step 3: Enthalpy for 6 g
\[ \Delta H = 0.2143 \times (-300) = -64.29\ \text{kJ} \]

Step 4: Check the options
Option (D) is double this value and (B) is half of it. Option (A) would result from dividing by 4 twice. Only \(-64.29\) kJ matches.

Final Answer:
The enthalpy change is -64.29 kJ. \[ \boxed{\text{(C)}\ -64.29\ \text{kJ}} \]