If 0.5 mol of H_2 and 0.5 mol of I_2 are re | vedclass

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

Question

(A) The reaction is: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$The equilibrium constant expression is: $K_c = \frac{[HI]^2}{[H_2][I_2]}$Since

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) The reaction is: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$The equilibrium constant expression is: $K_c = \frac{[HI]^2}{[H_2][I_2]}$Since the initial moles of $H_2$ and $I_2$ are equal ($0.5 \ mol$ each) in the same volume,their concentrations at equilibrium will also be equal,i.e.,$[H_2] = [I_2]$.Substituting this into the $K_c$ expression: $K_c = \frac{[HI]^2}{[I_2]^2}$Taking the square root on both sides: $\sqrt{K_c} = \frac{[HI]}{[I_2]}$Given $K_c = 49$,we have: $\frac{[HI]}{[I_2]} = \sqrt{49} = 7$.

If $0.5 \ mol$ of $H_2$ and $0.5 \ mol$ of $I_2$ are reacted in a $10 \ L$ vessel at $444 \ ^\circ C$,and the equilibrium constant $K_c$ at the same temperature is $49$,then the ratio of $[HI]$ to $[I_2]$ is .......

Similar Questions

The partial pressures of $CH_3OH$,$CO$ and $H_2$ in the equilibrium mixture for the reaction $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$ at $427 \ ^oC$ are $2.0 \ atm$,$1.0 \ atm$ and $0.1 \ atm$ respectively. The value of $K_P$ for the decomposition of $CH_3OH$ to $CO$ and $H_2$ is:

For which of the following reactions is $K_p = K_c$?

$x A_{(s)} \rightleftharpoons y B_{(g)} + z C_{(g)}$. If $\frac{K_c}{K_p} = (RT)^{-2}$,then which is correct?

The plot of $log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

The relation between equilibrium constant $K_p$ and $K_c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module