Question

Calculate \(\Delta H\) for a reaction, \(2X_{(g)}+Y_{(g)}\rightarrow 2Z_{(g)}\) at \(298\) K if \(\Delta U=-10.5\) kJ and \(R=8.314\ \text{J K}^{-1}\ \text{mol}^{-1}\).

• A. \(-14.45\) kJ
• B. \(-12.98\) kJ
• C. \(10.50\) kJ
• D. \(11.00\) kJ

Hint:

Use delta H = delta U + delta n(g) RT with the change in gaseous moles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a gas-phase reaction, \(\Delta H=\Delta U+\Delta n_g RT\).

Step 2: Find the Change in Moles
Products: 2 mol. Reactants: \(2+1=3\) mol. So \(\Delta n_g=2-3=-1\).

Step 3: Calculation
\(\Delta n_g RT=-1\times8.314\times298\) J \(=-2478\) J \(=-2.48\) kJ.

Step 4: Final Value
\(\Delta H=-10.5-2.48=-12.98\) kJ.

Final Answer:
The enthalpy change is \(-12.98\) kJ. \[ \boxed{\text{(B) }-12.98\ \text{kJ}} \]