$A$ reaction mixture containing $H_2, N_2$ and $NH_3$ has partial pressures of $2 \ atm, 1 \ atm$ and $3 \ atm$ respectively at $725 \ K.$ If the value of $K_P$ for the reaction,$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is $4.28 \times 10^{-5} \ atm^{-2}$ at $725 \ K,$ in which direction will the net reaction proceed?

$PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2} \quad K_{c} = 1.844$
$3.0 \ \text{moles}$ of $PCl_{5}$ is introduced in a $1 \ \text{L}$ closed reaction vessel at $380 \ \text{K}$. The number of moles of $PCl_{5}$ at equilibrium is $..... \times 10^{-3}$. (Round off to the Nearest Integer)

The reaction quotient $Q_{C}$ is useful in predicting the direction of the reaction. Which of the following is incorrect?

Consider a reaction that is first order in both directions: $A \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} B$. Initially only $A$ is present,and its concentration is $A_{0}$. Assume $A_{t}$ and $A_{\text{eq}}$ are the concentrations of $A$ at time $t$ and at equilibrium,respectively. The time $t$ at which $A_{t} = (A_{0} + A_{\text{eq}})/2$ is $....$

$2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2_{(g)}}$. If $NOBr$ is $40\%$ dissociated at a certain temperature and a total pressure of $0.30 \text{ atm}$,the $K_p$ for the reaction $2NO_{(g)} + Br_{2_{(g)}} \rightleftharpoons 2NOBr_{(g)}$ is:

The surface of copper gets tarnished by the formation of copper oxide. $N_2$ gas was passed to prevent the oxide formation during heating of copper at $1250 \ K$. However,the $N_2$ gas contains $1 \ \text{mole}\%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
$2 Cu_{(s)} + H_2O_{(g)} \longrightarrow Cu_2O_{(s)} + H_{2(g)}$
$p_{H_2}$ is the minimum partial pressure of $H_2$ (in $\text{bar}$) needed to prevent the oxidation at $1250 \ K$. The value of $\ln(p_{H_2})$ is . . . . .
(Given: total pressure $= 1 \ \text{bar}$,$R = 8 \ J \ K^{-1} \ mol^{-1}$,$\ln(10) = 2.3$. $Cu_{(s)}$ and $Cu_2O_{(s)}$ are mutually immiscible.
At $1250 \ K$: $2 Cu_{(s)} + 1/2 O_{2(g)} \longrightarrow Cu_2O_{(s)}; \Delta G^\theta = -78,000 \ J \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)}; \Delta G^\theta = -1,78,000 \ J \ mol^{-1}$)