Exams › JEE Main › Chemistry
ΔG° = −nFE° = ΔH° − TΔS° = (ΔH° + nFE°)/T = [−825.2×10^3 + (2×96487×4.315)] / 298 = 25.11 JK−1 mol−1 ∴ Nearest integer answer is 25
- 25
- 26
- 24
- 27
Correct answer: 25
Solution
The calculation correctly applies the thermodynamic relationship between Gibbs free energy, enthalpy, and entropy, leading to a value of 25.11 JK−1 mol−1, which rounds to the nearest integer of 25.
Related JEE Main Chemistry questions
- Using the following thermochemical equations, determine the enthalpy of vaporisation of water: \[ \mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \to H_2O(l)},\quad \Delta H = -286\ \text{kJ} \] \[ \mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \to H_2O(g)},\quad \Delta H = -245.5\ \text{kJ} \]
- Given that the standard enthalpy of formation of NH$_3$ is $-46.0\ \text{kJ mol}^{-1}$, and the enthalpy of formation of H$_2$ from gaseous atoms is $-436\ \text{kJ mol}^{-1}$ while that of N$_2$ from gaseous atoms is $-712\ \text{kJ mol}^{-1}$, what is the average N–H bond enthalpy in NH$_3$?
- Using the following bond dissociation data, determine the enthalpy change for the conversion $8S(s) + \tfrac32 S_8(g)$.
- Using the bond enthalpies given below, determine the enthalpy change for the hydrogenation reaction shown: H–H bond energy = 431.37 kJ mol$^{-1}$ C–C bond energy = 606.10 kJ mol$^{-1}$ C–H bond energy = 336.49 kJ mol$^{-1}$ C–H bond energy = 410.50 kJ mol$^{-1}$ $C=C + H–H \rightarrow H–C–C–H$ What is the enthalpy change for this reaction?
- For a given reversible reaction, both enthalpy change ($\Delta H$) and entropy change ($\Delta S$) are positive. If $T_1$ denotes the equilibrium temperature, under which condition will the reaction be spontaneous?
- For the spontaneous reaction $2C_2H_6(g) + 25O_2(g) \rightarrow 16CO_2(g) + 18H_2O(g)$, what are the signs of $\Delta H$, $\Delta S$, and $\Delta G$ in that order?
⚔️ Practice JEE Main Chemistry free + battle 1v1 →