Consider the reaction,$P(aq) \rightleftharpoons Q(aq)$ with an equilibrium constant $K=1.5$. The reaction is started in a vessel with a concentration of $[P]$ of $2 \ M$ and concentration of $[Q]=0$. When the equilibrium is established,half the amount of $P$ is removed,and the reaction is allowed to re-equilibrate. The concentration of $Q$ in the vessel (in $M$) is closest to

One mole each of $He$ and $A(g)$ are taken in a $10 \text{ L}$ closed flask and heated to $400 \text{ K}$ to establish the following equilibrium: $A(g) \rightleftharpoons B(g)$. $K_{c}$ for this reaction at $400 \text{ K}$ is $4.0$. The partial pressures (in $\text{atm}$) of $He$ and $B(g)$ are respectively (at equilibrium) (Assume $He$,$A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$)

$5 \ moles$ of $SO_2$ and $5 \ moles$ of $O_2$ are allowed to react. At equilibrium,it was found that $60\%$ of $SO_2$ is used up. If the partial pressure of the equilibrium mixture is $1 \ atm$,the partial pressure of $O_2$ is (in $atm$)

An equilibrium shifts towards reactants at higher temperatures. Find the correct graph.

The equilibrium constants of the following are
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,; \quad K_1$
$N_2 + O_2 \rightleftharpoons 2NO \,; \quad K_2$
$H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \,; \quad K_3$
The equilibrium constant $(K)$ of the reaction:
$2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O$ is:

The reaction $2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$ is at equilibrium in a closed $15 \ L$ vessel at $300 \ K$. The total weight of the mixture of $NO_2$ and $N_2O_4$ in the vessel is $64.4 \ g$. The equilibrium constant for the reaction is $K_p = 6.67$. Assuming ideal gas behavior,the total pressure in the vessel (in $atm$) is: [Given: Gas constant $R = 0.082 \ atm \ L \ K^{-1} \ mol^{-1}$]