The variation of equilibrium constant with temperature is given below:

$T_{1} = 25^{\circ}C$	$K_{1} = 100$
$T_{2} = 100^{\circ}C$	$K_{2} = 100$

The values of $\Delta H^{\circ}$,$\Delta G^{\circ}$ at $T_{1}$ and $\Delta G^{\circ}$ at $T_{2}$ (in $kJ \ mol^{-1}$) respectively,are close to: [Use $R = 8.314 \ J \ K^{-1} \ mol^{-1}$]

In a $2 \ L$ vessel,$1 \ mol$ of $H_2$ and $2 \ mol$ of $I_2$ are taken. If at equilibrium the moles of $H_2$ are $0.2$,then the moles of $I_2$ and $HI$ at equilibrium will be respectively:

One mole of $O_{2(g)}$ and two moles of $SO_{2(g)}$ were heated in a closed vessel of $1 \, L$ capacity at $1098 \, K$. At equilibrium,$1.6 \, moles$ of $SO_{3(g)}$ were found. The equilibrium constant $K_c$ of the reaction would be

The standard Gibbs energy change at $300 \, K$ for the reaction $2A \rightleftharpoons B + C$ is $2494.2 \, J$. At a given time,the composition of the reaction mixture is $[A] = 1/2, [B] = 2$ and $[C] = 1/2$. The reaction proceeds in the:

If the equilibrium constant $K_c = 0.04$,how many moles/liter of $PCl_5$ are required to obtain $0.1$ mole of $Cl_2$?

$K_{c}$ for the following reaction is $99.0$: $A_{2(g)} \rightleftharpoons B_{2(g)}$. In a $1 \ L$ flask,$2 \ moles$ of $A_{2}$ were heated to $T(K)$ and equilibrium was reached. The concentrations at equilibrium of $A_{2}$ and $B_{2}$ are $C_{1}(A_{2})$ and $C_{2}(B_{2})$ respectively. Now,$1 \ mole$ of $A_{2}$ was added to the flask and heated to $T(K)$ to establish equilibrium again. The concentrations of $A_{2}$ and $B_{2}$ are $C_{3}(A_{2})$ and $C_{4}(B_{2})$ respectively. What is the value of $C_{3}(A_{2})$ in $mol \ L^{-1}$?