3.00 , mol of PCl_5 kept in 1 , L closed vess | vedclass

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

Question

(A) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$At $t=0$: $[PCl_5] = 3 \, M$,$[PCl_3] = 0$,$[Cl_2] = 0$At equil

Vedclass · Accepted Answer

$1.59 \, M, 1.59 \, M$

Vedclass · Answer

(A) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$At $t=0$: $[PCl_5] = 3 \, M$,$[PCl_3] = 0$,$[Cl_2] = 0$At equilibrium: $[PCl_5] = (3-x) \, M$,$[PCl_3] = x \, M$,$[Cl_2] = x \, M$$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = 1.80$$1.80 = \frac{x^2}{3-x}$$5.4 - 1.8x = x^2$$x^2 + 1.8x - 5.4 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-1.8 + \sqrt{(1.8)^2 - 4(1)(-5.4)}}{2(1)}$$x = \frac{-1.8 + \sqrt{3.24 + 21.6}}{2} = \frac{-1.8 + \sqrt{24.84}}{2}$$x = \frac{-1.8 + 4.984}{2} = \frac{3.184}{2} \approx 1.59 \, M$Therefore,$[PCl_3] = 1.59 \, M$ and $[Cl_2] = 1.59 \, M$.

$3.00 \, mol$ of $PCl_5$ kept in $1 \, L$ closed vessel was allowed to attain equilibrium at $380 \, K$. If $K_C = 1.80$,calculate the equilibrium concentrations of $PCl_3$ and $Cl_2$ respectively.

Similar Questions

For the gaseous reaction $aA + bB \rightleftharpoons cC + dD$,the relation between $K_p$ and $K_c$ is:

For the following reaction in gaseous phase $CO + 1/2 O_2 \to CO_2$,the ratio $K_p/K_c$ is:

The degree of dissociation of $PCl_{5(g)}$ at $16.8 \ bar$ and $127 \ ^oC$ is $0.4$. The value of $K_p$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ at $25^o C$,the value of $K_c$ is $6.10 \times 10^{-3}$. What is the value of $K_c$ for the equilibrium reaction $NO_{2(g)} \rightleftharpoons 1/2 N_2O_{4(g)}$?

At $T(K)$,when $1 \ mol$ of $X$ and $1 \ mol$ of $Y$ are heated in a $1 \ L$ flask,$0.5 \ mol$ of $Z$ is formed at equilibrium. The $K_c$ value for the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)} + A_{(g)}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module