Calculate the change of entropy $Δ_{r} S_{m}^{\circ}$ at 298 K for the reaction in which urea is formed from $N H_{3}$ and $C O_{2}$ . The standard entropy of $N H_{2} C O N H_{2}$ is 174.0 $J K^{- 1} m o l^{- 1}$ . For $N H_{3}$ , $C O_{2}$ and $H_{2} O$ it is 192.3, 213 and 69.9 $J K^{- 1}$

- 354.4 J K^{- 1}

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354.4 J K^{- 1}

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- 453.4 J K^{- 1}

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453.4 J K^{- 1}

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Solution

The correct option is A $- 354.4 J K^{- 1}$
$Δ_{r} S_{m}^{\circ} = \sum V_{p} S^{\circ} (p r o d u c t s) - \sum V_{p} S^{\circ} (r e a c t a n t s) = [(174.0 + 69.9) - (2 \times 192.3 + 213)] J K^{- 1} m o l^{- 1} = (243.9 - 598.8) J K^{- 1} m o l^{- 1} = - 354.4 J K^{- 1}$

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Similar questions

Q. Calculate the change of entropy

Δ_{r} S_{m}^{\circ}

at 298 K for the reaction in which urea is formed from

N H_{3}

and

C O_{2}

. The standard entropy of

N H_{2} C O N H_{2}

is 174.0

J K^{- 1} m o l^{- 1}

. For

N H_{3}

C O_{2}

and

H_{2} O

it is 192.3, 213 and 69.9

J K^{- 1}

Standard enthalpy and standard entropy for the oxidation of $N H_{3}$ at $298$ K are $- 382.64$ $kJ {mol}^{- 1}$ and $- 145.6$ $J {mol}^{- 1} K^{- 1}$ respectively. Standard free energy change (in ${kJ mol}^{- 1}$ ) for the same reaction at $298$ K is: