Solveeit Logo
Login

Question

Chemistry Question on Gibbs Free Energy

For the given reaction, H2(g)+Cl2(g)−>2H+(aq)+2Cl−(aq);ΔG∘=−−262⋅4kJH _{2}( g )+ Cl _{2}( g )->2 H ^{+}( aq )+2 Cl ^{-}( aq ) ; \Delta G ^{\circ}=--262 \cdot 4 kJ The value of free energy of formation (ΔGf∘)\left(\Delta G_{f}^{\circ}\right) for the ion Cl−1(aq)C l^{-1}(a q), therefore will be

A

−131.2 kJ mol−1- 131.2 \, kJ \, mol^{-1}

B

+131.2 kJ mol−1 + 131.2 \, kJ \, mol^{-1}

C

−262.4 kJ mol−1- 262.4 \, kJ \, mol^{-1}

D

+262.4 kJ mol−1 + 262.4 \, kJ \, mol^{-1}

Answer

−131.2 kJ mol−1- 131.2 \, kJ \, mol^{-1}

Explanation

Solution

(ΔG∘)\left(\Delta G^{\circ}\right) reaction =ΔGf0= \Delta G^{0}_{f} (Products) −ΔGf0 - \Delta G^{0}_{f} (reactants)
∴264.4=[2ΔGf0(H+)+2ΔGf0(C1−)]\therefore 264.4 = \left[2 \Delta G^{0}_{f} \left(H^{+}\right) + 2 \Delta G^{0}_{f} \left(C^{1-}\right)\right]
or 264.4=−[ΔGf0(H2)+ΔGf0(Cl2)]264.4 = - \left[\Delta G_{f}^{0} \left(H_{2}\right) + \Delta G^{0}_{f} \left(Cl_{2}\right)\right]
=[0+2ΔGf0(Cl−)]+[0+0]= \left[0+ 2\Delta G^{0}_{f} \left(Cl^{-}\right)\right] + \left[0 +0\right]
or, −262.4=2ΔGf0(Cl−)- 262.4 = 2\Delta G_{f}^{0} \left(Cl^{-}\right)
or, ΔGg0(Cl−)=−131.2 kJ mol−1 \Delta G^{0}_{g}\left(Cl^{-}\right) = -131.2 \,kJ\,mol^{-1}