For a reversible reaction,the rate constants for the forward and backward reactions are $2.38 \times 10^{-4}$ and $8.15 \times 10^{-5}$ respectively. The equilibrium constant for the reaction is

The equilibrium constant $K_c$ for the reaction $HA + B \rightleftharpoons BH^{+} + A^{-}$ is $100$. If the rate constant for the forward reaction is $10^5$,then the rate constant for the backward reaction is:

$\Delta n$,the change in the number of moles for the reaction,$C_{12}H_{22}O_{11(s)} + 12O_{2(g)} \rightleftharpoons 12CO_{2(g)} + 11H_2O_{(l)}$ at $25 \ ^\circ C$ is

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

For the reaction system $A_{(g)} \rightleftharpoons B_{(g)} + 2C_{(g)}$,if the equilibrium concentration of $B$ is doubled,the equilibrium concentration of $C$ will become ....

The value of $\log \ K$ for the reaction $A \rightleftharpoons B$ at $298 \ K$ is (Nearest integer).
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)