For the equilibrium, {PCl}_{5( {g})} ⇌ {PCl}_{3}( {g})+ {Cl}_{2( {g})} at 298 {K}, K_{c}=1.8 \times 10^{-7}. Then. ∆ G for this reaction will be _______ [Given \log 4.5=0.6484]

Correct Answer is : 19.12kJ.mol−1

AP EAPCET 7 Sep 2021 Shift 1 Paper

Question : 134 of 160

Marks: +1, -0

For the equilibrium, PCl5(g)⇌PCl3(g)+Cl2(g) at 298K,Kc=1.8×10−7. Then. ∆G for this reaction will be _______ [Given log‌4.5=0.6484]

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