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

In a one-litre flask,$6$ moles of $A$ undergoes the reaction $A_{(g)} \rightleftharpoons P_{(g)}$. The progress of product formation at two temperatures (in Kelvin),$T_1$ and $T_2$,is shown in the figure:
If $T_1=2 T_2$ and $(\Delta G_2^{\Theta}-\Delta G_1^{\Theta})=R T_2 \ln x$,then the value of $x$ is. . . . .
$[\Delta G_1^{\Theta}$ and $\Delta G_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $T_1$ and $T_2$,respectively.]

For a given reaction,$2 A \rightleftharpoons B + C$,the equilibrium constant is $2 \times 10^{-3}$. If at any given time the composition of the reaction mixture is $[A]=[B]=[C]=6 \times 10^{-5} \ M$; predict in which direction the reaction will proceed and the correct value for reaction quotient.

The following reaction is performed at $298 \, K$.
$2 NO_{(g)} + O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$
The standard free energy of formation of $NO_{(g)}$ is $86.6 \, kJ/mol$ at $298 \, K$. What is the standard free energy of formation of $NO_{2(g)}$ at $298 \, K$? $(K_p = 1.6 \times 10^{12})$

For the equilibrium $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,the partial pressures of $SO_2$,$O_2$,and $SO_3$ are $0.662 \ atm$,$0.101 \ atm$,and $0.331 \ atm$ respectively. If the equilibrium concentrations of $SO_2$ and $SO_3$ are made equal,the partial pressure of $O_2$ will be ..... $atm$.

$28 \, g$ of $N_2$ and $6 \, g$ of $H_2$ were kept at $400 \, ^oC$ in a $1 \, L$ vessel. The equilibrium mixture contained $27.54 \, g$ of $NH_3$. The approximate value of $K_c$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is (in $L^2 \, mol^{-2}$):