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

At $300 \mathrm{~K}$ the enthalpy change for the following reaction is $-2800 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  1. $-2802.49$
  2. $-2800.00$
  3. $-2814.94$
  4. $+2802.49$

Solution

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

Asked in: AP EAMCET 2023 (15 May Shift 2)

Practice more Thermodynamics (C) questions on Aicharya