At T(K),the equilibrium constant of H_{2(g)} | vedclass

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(B) Given,the equilibrium constant $K_C = 49$. The equilibrium concentrations are $[H_2] = 2.0 	imes 10^{-2} \ M$ and $[I_2] = 8.0 	imes 10^{-2} \ M

Vedclass · Accepted Answer

$0.28$

Vedclass · Answer

(B) Given,the equilibrium constant $K_C = 49$. The equilibrium concentrations are $[H_2] = 2.0 	imes 10^{-2} \ M$ and $[I_2] = 8.0 	imes 10^{-2} \ M$. The expression for the equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} ightleftharpoons 2 HI_{(g)}$ is $K_C = \frac{[HI]^2}{[H_2][I_2]}$. Rearranging the formula to solve for $[HI]$,we get $[HI]^2 = K_C 	imes [H_2] 	imes [I_2]$. Substituting the given values: $[HI]^2 = 49 	imes (2.0 	imes 10^{-2}) 	imes (8.0 	imes 10^{-2}) = 49 	imes 16 	imes 10^{-4}$. Taking the square root of both sides: $[HI] = \sqrt{49 	imes 16 	imes 10^{-4}} = 7 	imes 4 	imes 10^{-2} = 28 	imes 10^{-2} = 0.28 \ mol \ L^{-1}$.

At $T(K)$,the equilibrium constant of $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $49$. If $[H_2]$ and $[I_2]$ at equilibrium at the same temperature are $2.0 \times 10^{-2} \ M$ and $8.0 \times 10^{-2} \ M$ respectively,the $[HI]$ at equilibrium in $mol \ L^{-1}$ is:

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