For CaCO3(s) -> CaO(s) + CO2(g) take Delta H deg = +178 kJ mol^-1 and Delta S deg = +161 J mol^-1K^-1 (temperature independent). Find Delta G deg at 298 K in kJ/mol (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3td195

For $CaCO X_{3} (s) CaO (s) + CO X_{2} (g)$ take $Δ H^{\circ} = + 178 kJ mo l^{- 1}$ and $Δ S^{\circ} = + 161 J mo l^{- 1} K^{- 1}$ (temperature independent). Find $Δ G^{\circ}$ at $298 K$ in kJ/mol (two decimals).

Numerical answer type, answer in kJ/mol, tolerance 0.01.

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = 178 - 298 \times 0.161 = 178 - 47.978 = 130.02 kJ mo l^{- 1}