The enthalpy and entropy change for the reaction $Br_{2(l)} + Cl_{2(g)} \rightarrow 2BrCl_{(g)}$ are $30 \ kJ \ mol^{-1}$ and $105 \ J \ K^{-1} \ mol^{-1}$ respectively. The temperature at which the reaction will be in equilibrium is ............... $K$.

Calculate the work done (in $J$) when $4.5 \ g$ of $H_2O_2$ reacts against a pressure of $1.0 \ atm$ at $25 \ ^oC$. [$2H_2O_{2(l)} \to O_{2(g)} + 2H_2O_{(l)}$]

Which of the following is correct?

For the reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$,$\Delta H = -285.8 \ kJ \ mol^{-1}$,$\Delta S = -0.163 \ kJ \ mol^{-1} K^{-1}$. What is the value of free energy change at $27 \ ^\circ C$ for the reaction in $kJ \ mol^{-1}$?

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?

The standard state Gibbs free energies of formation of $C$ (graphite) and $C$ (diamond) at $T = 298 \ K$ are:
$\Delta_f G^0[C(\text{graphite})] = 0 \ kJ \ mol^{-1}$
$\Delta_f G^0[C(\text{diamond})] = 2.9 \ kJ \ mol^{-1}$
The standard state means that the pressure should be $1 \ bar$,and the substance should be pure at a given temperature. The conversion of graphite [$C$ (graphite)] to diamond [$C$ (diamond)] reduces its volume by $2 \times 10^{-6} \ m^3 \ mol^{-1}$. If $C$ (graphite) is converted to $C$ (diamond) isothermally at $T = 298 \ K$,the pressure at which $C$ (graphite) is in equilibrium with $C$ (diamond) is:
[Useful information: $1 \ J = 1 \ kg \ m^2 \ s^{-2} ; 1 \ Pa = 1 \ kg \ m^{-1} \ s^{-2} ; 1 \ bar = 10^5 \ Pa$ ] (in $bar$)