Consider the equilibrium,H_2 + I_2 rightlefth | vedclass

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Question

(B) The given reaction is $H_2 + I_2 \rightleftharpoons 2 HI$. The reverse reaction is $2 HI \rightleftharpoons H_2 + I_2$. The equilibrium constant f

Vedclass · Accepted Answer

$0.021$

Vedclass · Answer

(B) The given reaction is $H_2 + I_2 ightleftharpoons 2 HI$. The reverse reaction is $2 HI ightleftharpoons H_2 + I_2$. The equilibrium constant for the reverse reaction is given by $K = \frac{[H_2][I_2]}{[HI]^2}$. Given concentrations are $[H_2] = 1.14 	imes 10^{-2} \ mol \ L^{-1}$,$[I_2] = 0.12 	imes 10^{-2} \ mol \ L^{-1}$,and $[HI] = 2.52 	imes 10^{-2} \ mol \ L^{-1}$. Substituting these values into the expression: $K = \frac{(1.14 	imes 10^{-2}) 	imes (0.12 	imes 10^{-2})}{(2.52 	imes 10^{-2})^2}$ $K = \frac{0.1368 	imes 10^{-4}}{6.3504 	imes 10^{-4}}$ $K \approx 0.02154$. Rounding to two significant figures,$K = 0.021$.

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