At a definite temperature and $3 \ atm$ pressure,$75\%$ of $PCl_5$ decomposes into $PCl_3$ and $Cl_2$. Find $K_p$ for the reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ (in $atm$)

  • A
    $1.85$
  • B
    $3.857$
  • C
    $2.50$
  • D
    $4.20$

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Similar Questions

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

In the synthesis of $HI$,the amounts of $H_{2(g)}$,$I_{2(g)}$,and $HI_{(g)}$ at equilibrium were found to be $0.8 \ mol$,$0.8 \ mol$,and $2.4 \ mol$ respectively in a $10 \ L$ vessel. Calculate the equilibrium constant $(K_c)$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ and the equilibrium constant for the reverse reaction.

$A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$. The correct relationship between $K_P$,$\alpha$,and equilibrium pressure $P$ is:

$18.4 \ g$ of $N_2O_4$ was placed in a $1 \ L$ vessel at $400 \ K$ and allowed to attain the following equilibrium: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. If the total pressure at equilibrium was $10.64 \ bar$,the approximate $K_p$ is (Given: $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$,assume $N_2O_4$ and $NO_2$ behave as ideal gases).

The equilibrium constant $K_p$ for the following reaction at $191\,^{\circ}C$ is $1.24$. What is the value of $K_c$? $B_{(s)} + \frac{3}{2}F_{2(g)} \rightleftharpoons BF_{3(g)}$

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