The enthalpy change for the reaction ${CaCO}_{3} (s) \to CaO (s) + {CO}_{2} (g)$ at $1000 K$ is $176 k J {mol}^{- 1}$ . The value of $Δ U$ for the reaction will be

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$176 kJ {mol}^{- 1}$

$180 kJ {mol}^{- 1}$

$167.7 kJ {mol}^{- 1}$

$184.3 kJ {mol}^{- 1}$

answer is C.

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Detailed Solution

$Δ v_{g} = 1 .$

Hence,

$\begin{array}{l} Δ U = Δ H - (Δ v_{g}) R T = 176 kJ {mol}^{- 1} - (1) (8.314 \times 10^{- 3} {kJK}^{- 1} {mol}^{- 1}) (1000 K) \\ = (176 - 8.314) kJ {mol}^{- 1} = 167.7 kJ {mol}^{- 1} \end{array}$

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