Question

The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are 382.64 kJ mol${}^{-1}$ and 145.6 J K${}^{-1}$ mol${}^{-1},$ respectively. Standard Gibb's energy change for the same reaction at 298 K is:

• A. -2221.1 kJ mol${}^{-1}$
• B. -339.3 kJ mol${}^{-1}$
• C. -439.3 kJ mol${}^{-1}$
• D. -523.2 kJ mol${}^{-1}$

Answer 1

The correct option is B -339.3 kJ mol${}^{-1}$

Expression for Gibb's free energy is given by,

$\Delta G^0=\Delta H-T.\Delta S^0$ ........(1)

Given that

$T=$ 298 $K$

$\Delta H=-382.64kJmo{l}^{-1}$

$\Delta S^0=-145.6J{K}^{-1}mo{l}^{-1}$

$=-145.6\times {10}^{-3}kJ{K}^{-1}$

On substituting these values in equation (1), we get,

$\Delta G^0=-382.64-(298\times -145.6\times {10}^{-3})$

$=-339.3kJmo{l}^{-1}$

Hence, the correct option is $B$