For the reaction H_{2(g)} + I_{2(g)} rightlef | vedclass

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

Question

(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Initial moles: $H_2 = 0.4$,$I_2 = 0.4$,$HI = 0$.At equilibrium,$n(HI) = 0.5 \ m

Vedclass · Accepted Answer

$11.11$

Vedclass · Answer

(A) The reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Initial moles: $H_2 = 0.4$,$I_2 = 0.4$,$HI = 0$.At equilibrium,$n(HI) = 0.5 \ mol$.From stoichiometry,$2 \ mol$ of $HI$ are formed from $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.So,$0.5 \ mol$ of $HI$ is formed from $0.25 \ mol$ of $H_2$ and $0.25 \ mol$ of $I_2$.Equilibrium moles: $n(H_2) = 0.4 - 0.25 = 0.15 \ mol$,$n(I_2) = 0.4 - 0.25 = 0.15 \ mol$,$n(HI) = 0.5 \ mol$.Concentrations in $2 \ L$ vessel: $[H_2] = 0.15/2 = 0.075 \ M$,$[I_2] = 0.075 \ M$,$[HI] = 0.5/2 = 0.25 \ M$.$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.25)^2}{(0.075)(0.075)} = \frac{0.0625}{0.005625} = 11.11$.Since $\Delta n_g = 2 - (1+1) = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c = 11.11$.

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$0.4 \ mol$ of $H_2$ and $I_2$ each are taken in a $2 \ L$ vessel. If $0.5 \ mol$ of $HI$ is formed at equilibrium,calculate the equilibrium constant $K_c$ and $K_p$.

Similar Questions

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the value of $K_c$ is $1.8 \times 10^{-6}$ at $184^{\circ}C$. Given $R = 0.083 \ L \cdot bar \cdot K^{-1} \cdot mol^{-1}$,compare $K_p$ and $K_c$ at $184^{\circ}C$.

For the reaction $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$,if the reaction is started with $1 \ mol$ of $H_2$ and $CO_2$ each,and $x \ mol/L$ of $H_2$ is consumed at equilibrium,then the correct expression for $K_p$ will be ........

For the reaction $H_2 + I_2 \rightleftharpoons 2HI$,the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $8.0 \ mol \ L^{-1}$,$3.0 \ mol \ L^{-1}$,and $28.0 \ mol \ L^{-1}$ respectively. The equilibrium constant $(K_c)$ of the reaction is: (in $.66$)

For the reaction,$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_c$ at $250 \ ^oC$ is $26$. The value of $K_p$ at this temperature will be:

For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module