The value of $\log \ K$ for the reaction $A \rightleftharpoons B$ at $298 \ K$ is (Nearest integer).
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)

$2X_{6(g)} \rightleftharpoons 4X_{3(g)}$
$A$ $20 \ L$ box contains $X_6$ and $X_3$ at equilibrium at $1500 \ K$. The equilibrium constant $K_p = 4 \times 10^{18} \ atm$. Assuming $P_{X_3} \gg P_{X_6}$ and the total pressure is $10 \ atm$,the partial pressure of $X_6$ will be:

$40\%$ of $HI$ undergoes decomposition to $H_2$ and $I_2$ at $300 \ K$. $\Delta G^{\ominus}$ for this decomposition reaction at one atmosphere pressure is $... \ J \ mol^{-1}$. [nearest integer]
(Use $R = 8.31 \ J \ K^{-1} \ mol^{-1}$; $\log 2 = 0.3010$; $\ln 10 = 2.3$; $\log 3 = 0.477$)

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

For the reaction $A_{(g)} \rightleftharpoons 2 B_{(g)}$,the backward reaction rate constant is higher than the forward reaction rate constant by a factor of $2500$,at $1000 \ K$. [Given : $R = 0.0831 \ L \ atm \ mol^{-1} \ K^{-1}$] $K_p$ for the reaction at $1000 \ K$ is

For $NH_4HS_{(s)} \rightleftharpoons NH_{3_{(g)}} + H_2S_{(g)}$; the observed pressure for the reaction mixture at equilibrium is $1.12 \ atm$ at $106 \ ^oC$. The value of $K_p$ for the reaction is: (in $atm^2$)