For the reaction A(g)+B(g)→C(g)+D(g) ΔH∘ and ΔS∘ are, respectively, −29.8kJmol−1 and −0.100kJK−1mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is:

# For the reaction  are, respectively, $-29.8\mathrm{kJmol}-1$ and $-0.100{\mathrm{kJK}}^{-1}{\mathrm{mol}}^{-1}$ at 298 K. The equilibrium constant for the reaction at 298 K is:

1. A

$1.0×{10}^{-10}$

2. B

10

3. C

1

4. D

$1.0×{10}^{10}$

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### Solution:

$\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }=\left(-29.8\mathrm{kJ}{\mathrm{mol}}^{-1}\right)-\left(298\mathrm{K}\right)\left(-0.100{\mathrm{kJK}}^{-1}{\mathrm{mol}}^{-1}\right)=0$

From the expression, $\mathrm{\Delta }{G}^{\circ }=-RT\mathrm{ln}{K}_{p}^{\circ }$, we get ${K}_{p}^{\circ }=1$

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