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Q.

For the reaction taking place at constant pressure $X_{(g)} ⇌ Y_{(g)}$ the gibbs free energy change at $27^{0} C$ and $52^{0} C$ is -80KJ/mole and -30 KJ/mole respectively. Identify the correctly drawn conclusions. (All the symbols have usual meaning given in the thermodynamics)

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a

The enthalpy change of reaction at $52^{0} C$ is -930KJ/mole

b

${(Δ_{r} S^{0})}_{P}^{}$ from $27^{0} C t o 52^{0} C$ is -2KJ/mole/K

c

The reaction is exothermic at $52^{0} C$

d

${(\frac{d (Δ_{r} G^{0})}{d T})}_{p}$ from $27^{0} C t o 52^{0} C$ is 2KJ/mole/K

answer is A, C, D.

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Detailed Solution

$Δ G = Δ H + T {[\frac{d (Δ G)}{d T}]}_{p} \frac{d (Δ G)}{d T} = \frac{[- 30] - [- 80]}{325 - 300} = \frac{- 30 + 80}{25} = \frac{50}{25} = 2 Δ G = Δ H + T {[\frac{d Δ G}{d T}]}_{p} - 30 = Δ H + 325 [2] Δ H = - 30 - 650 = - 680 K J / m o l e$

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