Start free

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 23 January, Morning Shift — Question 48

The standard enthalpy and standard entropy of decomposition of N2O4\mathrm{N}_{2} \mathrm{O}_{4} to NO2\mathrm{NO}_{2} are 55.0 kJ mol−155.0 \mathrm{~kJ} \mathrm{~mol}^{-1} and 175.0 J/K/mol175.0 \mathrm{~J} / \mathrm{K} / \mathrm{mol} respectively. The standard free energy change for this reaction at 25∘C25^{\circ} \mathrm{C} in Jmol−1\mathrm{J} \mathrm{mol}^{-1} is____ (Nearest integer)

Answer: 2850

Numerical answer — enter this value.

Step-by-step solution

ΔHrxno=55 kJ/mol,T=298 K\Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}=55 \mathrm{~kJ} / \mathrm{mol}, \quad \mathrm{T}=298 \mathrm{~K}

ΔSrxn0=175 J/mol\Delta \mathrm{S}_{\mathrm{rxn}}^{0}=175 \mathrm{~J} / \mathrm{mol}

ΔGrxno=ΔHrxno−TΔSrxxo\Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{rxx}}^{\mathrm{o}}

⇒ΔGrxn0=55000 J/mol−298×175 J/mol\Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{0}=55000 \mathrm{~J} / \mathrm{mol}-298 \times 175 \mathrm{~J} / \mathrm{mol}

⇒ΔGrxno=55000−52150\Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000-52150

⇒ΔGrxno=2850 J/mol\Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=2850 \mathrm{~J} / \mathrm{mol}

Answer key and solution verified before publishing.

Practise Thermodynamics & Thermochemistry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
23 January, Morning Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry

More Thermodynamics & Thermochemistry questions from this paper

The standard enthalpy and standard entropy of decomposition of N 2 O… | JEE Main 2025 PYQ with Solution · DhiX AI