Consider the equilibrium,H_2 + I_2 rightlefth | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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,$K = 0.021$.

Similar Questions

At $380 \ K$,$NH_4HS$ decomposes and the total pressure is $1.12 \ bar$. Find $K_p$ for the reaction: $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$ (in $bar^2$)

The value of $K_P / K_C$ for the reaction at $T(K)$ is:
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,$K_c = 1.8 \times 10^{-6}$ at $184 \, ^\circ C$. Given $R = 0.0831 \, kJ/(mol \cdot K)$,when $K_p$ and $K_c$ are compared at $184 \, ^\circ C$,it is found that:

Vedclass Test Series

Exam Paper Generator

Online Exam Module