In a closed container of 1000, cm^3,2, mol of | vedclass

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

Question

(D) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Given volume $V = 1000\, cm^3 = 1\, L$.The equilibrium concentr

Vedclass · Accepted Answer

$74$

Vedclass · Answer

(D) The equilibrium reaction is: $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Given volume $V = 1000\, cm^3 = 1\, L$.The equilibrium concentrations are:$[PCl_5] = 2\, mol / 1\, L = 2\, M$$[PCl_3] = 2\, mol / 1\, L = 2\, M$$[Cl_2] = 3\, mol / 1\, L = 3\, M$Calculate $K_C$:$K_C = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{2 	imes 3}{2} = 3$Calculate $K_P$ using the relation $K_P = K_C(RT)^{\Delta n_g}$:Here,$\Delta n_g = (1 + 1) - 1 = 1$,$R = 0.0821\, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 27 + 273 = 300\, K$.$K_P = 3 	imes (0.0821 	imes 300)^1 = 3 	imes 24.63 = 73.89 \approx 74\, atm$.

In a closed container of $1000\, cm^3$,$2\, mol$ of $PCl_5$,$2\, mol$ of $PCl_3$,and $3\, mol$ of $Cl_2$ are found to be at equilibrium at $27\, ^oC$. Then $K_P$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at $27\, ^oC$ is $.....$ $atm$.

Similar Questions

The equilibrium concentrations of $X$,$Y$,and $YX_2$ are $4 \ mol/L$,$2 \ mol/L$,and $2 \ mol/L$ respectively for the equilibrium $2X + Y \rightleftharpoons YX_2$. The value of $K_c$ is:

$8 \ mol$ of $AB_3$ is taken in a $1.0 \ dm^3$ container. It dissociates according to the reaction $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$. If $2 \ mol$ of $A_2$ are present at equilibrium,the equilibrium constant $K_c$ of the reaction is ....... $mol^2 \ L^{-2}$.

For the reaction $AB_{(g)} \rightleftharpoons A_{(g)} + B_{(g)}$,$AB$ is $33.3\%$ dissociated at a total equilibrium pressure of $P$. Therefore,$P$ is correctly related to $K_p$ by which of the following options?

For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the value of $K_p$ at $800 \ ^oC$ is $1.16 \ atm$. If $1 \ mol$ of $CaCO_{3(s)}$ is taken in a $1 \ L$ container to start the reaction,what will be the partial pressure of $CO_2$ at equilibrium in $atm$?

For a reversible reaction $A \rightleftharpoons B$,the equilibrium concentration of $B$,denoted as $[B]_e$,is given by which expression?

Vedclass Test Series

Exam Paper Generator

Online Exam Module