In a chemical equilibrium,the rate constant of the backward reaction is $7.5 \times 10^{-4}$ and the equilibrium constant is $1.5$. The rate constant of the forward reaction is:

What is the value of the equilibrium constant $K_p$ for the reaction ${H_2O_{(l)}} \rightleftharpoons {H_2O_{(g)}}$ at $60^{\circ}C$? (The vapor pressure of water at $60^{\circ}C$ is $0.185 \ bar$.)

At $673 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_c = 0.50$. Calculate $K_p$ at this temperature. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

For the reaction $2AB_{(g)} \rightleftharpoons 2A_{(g)} + B_{2(g)}$,$AB$ dissociates. If the initial pressure of $AB$ is $500 \, mm$ and the total pressure at equilibrium is $625 \, mm$,calculate $K_p$ for the reaction. Assume constant volume.

Equilibrium constants for the following reactions at $1200 \ K$ are given:
$2 \ H_2O_{(g)} \rightleftharpoons 2 \ H_{2(g)} + O_{2(g)}$
$K_1 = 6.4 \times 10^{-8}$
$2 \ CO_{2(g)} \rightleftharpoons 2 \ CO_{(g)} + O_{2(g)}$
$K_2 = 1.6 \times 10^{-6}$
The equilibrium constant for the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$
at $1200 \ K$ will be

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 $atm$.