Standard entropies of $X_2$,$Y_2$ and $XY_3$ are $60$,$40$ and $50 \ J \ K^{-1} \ mol^{-1}$ respectively. For the reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightarrow XY_3$,the enthalpy change is $\Delta H = -30 \ kJ \ mol^{-1}$. At what temperature will the reaction be at equilibrium (in $K$)?

When $1 \ g$ equivalent of a strong acid reacts with a strong base,the heat released is $13.5 \ kcal$. When $1 \ g$ equivalent of $H_2A$ is completely neutralized against a strong base,$13 \ kcal$ is released. When $1 \ g$ equivalent of $B(OH)_2$ is completely neutralized against a strong acid,$10 \ kcal$ of heat is released. Calculate the enthalpy change when $1 \ mole$ of $H_2A$ is completely neutralized by $B(OH)_2$ in $kcal$.

For the reaction,$2Cl_{(g)} \to Cl_{2(g)}$,the signs of $\Delta H$ and $\Delta S$ respectively,are:

Given below are two statements:
Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$
Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$
The correct answer is

Find the enthalpy of neutralisation in $kJ/mol$ for the reaction between $NH_4OH$ and $HCN$ in aqueous solution,given that the enthalpies of ionisation of $NH_4OH$ and $HCN$ are $7 \ kJ/mol$ and $8 \ kJ/mol$ respectively,and the enthalpy of neutralisation of a strong acid and a strong base is $-57.3 \ kJ/mol$.

$4 \ g$ of $NH_4NO_3$ were dissolved in $196 \ g$ of water at constant pressure in a calorimeter with a heat capacity of $160 \ J \ K^{-1}$. The temperature fell by $1.3 \ K$. The specific heat of the solution is $4.2 \ J \ K^{-1} \ g^{-1}$. The enthalpy of solution of $NH_4NO_3$ is $......$ $kJ \ mol^{-1}$.