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Q.

The equilibrium constant Kc for the following reaction at 842^oC is 7.90 x 10^-3.What is K _p at same temperature? $\frac{1}{2} F_{2} (g) ⇌ F (g)$

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a

$8.26 \times 10^{- 4}$

b

$7.90 \times 10^{- 2}$

c

$7.56 \times 10^{- 2}$

d

$8.64 \times 10^{- 5}$

answer is D.

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Detailed Solution

$k_{p} = k_{c} \cdot (RT)^{∆ n_{g}}$

$∆ n_{g} = 1 - \frac{1}{2} = \frac{1}{2}; T = 842 ° c = 1115 k K_{p} = 7.90 \times 10^{- 3} {(0.0821 \times 1115)}^{1 / 2} = 7.56 \times 10^{- 2}$

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The equilibrium constant Kc for the following reaction at 842oC is 7.90 x 10-3.What is K p at same temperature? 12 F2(g)⇌F(g)