For the reaction 2HI(g) ⇌H(g) + I2(g) at 250C, ∆G°=+1.7Kj Equilibrium constant for the reaction is

R = 8.314 J K-1mole-1T = 298 K∆G°=1.7×103 j = - 2.303RT logK1.7×103 = - 2.303(8.314)(298)logKlogK = - 0.3K = 0.5

For the reaction 2HI(g) ⇌H(g) + I2(g) at 250C, ∆G°=+1.7Kj Equilibrium constant for the reaction is

R = 8.314 J K-1mole-1T = 298 K∆G°=1.7×103 j = - 2.303RT logK1.7×103 = - 2.303(8.314)(298)logKlogK = - 0.3K = 0.5

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For the reaction 2HI_(g) $⇌$ H_(g) + I_2(g)at 25⁰C, $∆ G °$ =+1.7Kj Equilibrium constant for the reaction is

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Detailed Solution

R = 8.314 J K^-1mole^-1

T = 298 K

$∆ G ° =$ 1.7 $\times$ 10³ j = - 2.303RT logK

1.7 $\times$ 10³ = - 2.303(8.314)(298)logK

logK = - 0.3

K = 0.5

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For the reaction 2HI(g) ⇌H(g) + I2(g) at 250C, ∆G°=+1.7Kj Equilibrium constant for the reaction is

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For the reaction 2HI_(g) $⇌$ H_(g) + I_2(g)at 25⁰C, $∆ G °$ =+1.7Kj Equilibrium constant for the reaction is