$2 \ mol$ of $N_2O_{4(g)}$ is kept in a closed container at $298 \ K$ and under $1 \ atm$ pressure. It is heated to $596 \ K$ when $20 \%$ by mass of $N_2O_{4(g)}$ decomposes to $NO_2$. The resulting pressure is (in $atm$)

$K_{c}$ 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 $mol \ L^{-1}$?

For the reaction,$NO_2 + CO \rightleftharpoons NO + CO_2$,one mole of $NO_2$ and $2$ moles of $CO$ were kept in a vessel. Calculate the equilibrium constant $K_p$,if at equilibrium $25 \%$ of the initial amount of $CO$ is consumed.

$A$ plot of $\ln \ K$ against $\frac{1}{T}$ (abscissa) is expected to be a straight line with an intercept on the ordinate axis equal to:

Which of the following statements is/are wrong?
$(a)$ At equilibrium,concentrations of reactants and products become constant because the reaction stops.
$(b)$ Addition of catalyst speeds up the forward reaction more than the backward reaction.
$(c)$ Equilibrium constant of an exothermic reaction decreases with increase of temperature.
$(d)$ $K_p$ is always greater than $K_c$.

Consider a reversible reaction $2A + B \rightleftharpoons 2C$ having equilibrium constant $K_c = 25$. If a reaction vessel contains $2 \ mol$ of $A$,$0.25 \ mol$ of $B$,and $0.5 \ mol$ of $C$ in a $100 \ L$ vessel,what will be the direction of the reaction?