Pure PCl_5 is introduced into an evacuated ch | vedclass

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

Question

(C) The reaction is $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.Let the initial pressure of $PCl_5$ be $P_0$.At equilibrium,let the degree

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(C) The reaction is $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.Let the initial pressure of $PCl_5$ be $P_0$.At equilibrium,let the degree of dissociation be $\alpha$.The total pressure is $P_{total} = 2.0\ atm$.The equilibrium mixture contains $40\%$ $Cl_2$ by volume,which means the mole fraction of $Cl_2$ is $0.4$.Partial pressure of $Cl_2$ is $P_{Cl_2} = 0.4 	imes 2.0 = 0.8\ atm$.Since the stoichiometry is $1:1:1$,$P_{PCl_3} = P_{Cl_2} = 0.8\ atm$.The partial pressure of $PCl_5$ is $P_{PCl_5} = P_{total} - (P_{PCl_3} + P_{Cl_2}) = 2.0 - (0.8 + 0.8) = 0.4\ atm$.$K_p = \frac{P_{PCl_3} 	imes P_{Cl_2}}{P_{PCl_5}} = \frac{0.8 	imes 0.8}{0.4} = \frac{0.64}{0.4} = 1.6\ atm$.

$(1)$ $A_{2(g)} + 3B_{2(g)} \rightleftharpoons 2AB_{3(g)}$	$(i)$ $(RT)^{-2}$
$(2)$ $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$	$(ii)$ $(RT)^0$
$(3)$ $A_{(s)} + 1.5B_{2(g)} \rightleftharpoons AB_{3(g)}$	$(iii)$ $(RT)^{1/2}$
$(4)$ $AB_{2(g)} \rightleftharpoons AB_{(g)} + 0.5B_{2(g)}$	$(iv)$ $(RT)^{-1/2}$

Pure $PCl_5$ is introduced into an evacuated chamber and comes to equilibrium at $247\, ^oC$ and $2.0\ atm$. The equilibrium gaseous mixture contains $40\%$ chlorine by volume. Calculate $K_p$ at $247\, ^oC$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ in $atm$.

Similar Questions

For the formation of ammonia from its constituent elements ($1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$) in a closed vessel of volume $V \ L$,the value of $K_C$ is [units of $K_C = mol^{-2} \ L^2$].

For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,at equilibrium,the number of moles of $PCl_5$,$PCl_3$,and $Cl_2$ are $2 \ mol$ each,and the total pressure is $3 \ atm$. What is the value of $K_p$ in $atm$?

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 $.....$ $atm$.

$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$
$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$
Then the equilibrium constant for the following reaction is ....
$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module