Q11 / 160 · AP EAMCET 2023 (15 May Shift 2) · Chemistry

At $300 \mathrm{~K}$ the enthalpy change for the following reaction is $-2800 \mathrm{~kJ} \mathrm{~mol}^{-1}$

  • A
    $-2802.49$
  • Correct answer
    B
    $-2800.00$
  • C
    $-2814.94$
  • D
    $+2802.49$

Solution

$\begin{aligned} & \text {} \Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ & \Rightarrow \Delta \mathrm{U}=\Delta \mathrm{H}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}=(-2800)-(0 \times 0.082 \times 300) \\ & =-2800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \end{aligned}$

Hence, the correct answer is option: 'B'.

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