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

$XY_2$ dissociates as $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$. When the initial pressure of $XY_2$ is $600 \ mm \ Hg$,the total equilibrium pressure is $800 \ mm \ Hg$. Calculate $K_p$ for the reaction,assuming the volume of the system remains unchanged.

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

For the reaction $2AB_{(g)} \rightleftharpoons 2A_{(g)} + B_{2(g)}$; if the initial pressure of $AB$ is $100 \ atm$ and at equilibrium the total pressure becomes $125 \ atm$,the equilibrium constant $(K_p)$ will be:

The value of $K_{C}$ for the equilibrium reaction: $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ at $T \ K$ is $0.036$. If the equilibrium concentration of $CO_{2(g)}$ is $0.004 \ M$,the concentration of $CO_{(g)}$ in $mol \ L^{-1}$ is:

$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be