Verified Solution • 2025–26 Session moderate

QJEEM15Ch8ee0: JEE Main 2015 Chemistry - Class 11 JEEM15Ch8ee0 Solution: The following reaction is performed at 298 K 2NO(g) + O 2 (g)…

The following reaction is performed at 298 K
2NO(g) + O2 (g) ⇋\leftrightharpoons 2NO2 (g)
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K? (KP = 1.6 × 1012)

Multiple Choice Options

  • A 86600 + R(298) ln(1.6 ×\times 1012)
  • B 86600 - ln(1.6×1012)R(298)ln (1.6 \times 10^{12}) \over R (298)
  • C 0.5[2×86,600 – R(298) ln(1.6×1012)]
  • D R(298) ln(1.6×1012) – 86600

Step-by-Step Verified Solution

ΔG∘NO(g)\Delta {G^ \circ }_{NO\left( g \right)}

             =86.6kJ/mol=86600J/mol;\,\,\,\,\,\,\,\,\,\,\,\,\, = 86.6kJ/mol = 86600J/mol;

               \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, G∘NO2(g)=x J/mol{G^ \circ }_{N{O_2}\left( g \right)} = x\,J/mol

T=298, Kp=1.6×1012T = 298,\,{K_p} = 1.6 \times {10^{12}}

ΔG∘=−RT ln⁡ Kp\Delta {G^ \circ } = - RT\,\ln \,{K_p}

Given equation,

2NO(g)+O2(g) ⇌ 2NO2(g)2NO\left( g \right) + {O_2}\left( g \right)\,\rightleftharpoons\,2N{O_2}\left( g \right)

2ΔG∘NO2−2ΔG∘NO2\Delta {G^ \circ }_{N{O_2}} - 2\Delta {G^ \circ }_{NO}

               =−R(298)ln⁡(1.6×1012)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)

2ΔG∘NO2−2×866002\Delta {G^ \circ }_{N{O_2}} - 2 \times 86600

               =−R(298)ln⁡(1.6×1012)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)

2ΔG∘NO22\Delta {G^ \circ }_{N{O_2}}

              \,\,\,\,\,\,\,\,\,\,\,\,\,\, =2×86600−R(298)ln⁡(1.6×1012)= 2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)

ΔG∘NO2\Delta {G^ \circ }_{N{O_2}}

              \,\,\,\,\,\,\,\,\,\,\,\,\,\, =12[2×86600−R(298)ln⁡(1.6×1012)]= {1 \over 2}\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]

              \,\,\,\,\,\,\,\,\,\,\,\,\,\, =0.5[2×86600−R(298)ln⁡(1.6×1012)]= 0.5\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring