Dear Student, ∆G°=-2 303RTlog KeqHere, For 1 mol of HI(g),∆G°=107 kJ=107×103 JTherefore, ∆G°=107×103 Jmol-1T= 25°C= 25+273=298 KTherefore, 107×103 Jmol-1 = -2 303 (8 314 JK-1mol-1) (298 K) log Keqor, log Keq=-107×1032 303×8 314×298or, log Keq=-0 01875×103or, Keq= Antilog (-18 75) =1 77×10-19

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Dear Student,

∆ G ° = - 2.303 RTlog K_{eq} Here, For 1 mol of HI (g), ∆ G ° = 107 kJ = 107 \times 10^{3} J Therefore, ∆ G ° = 107 \times 10^{3} {Jmol}^{- 1} T = 25 ° C = 25 + 273 = 298 K Therefore, 107 \times 10^{3} {Jmol}^{- 1} = - 2.303 (8.314 {JK}^{- 1} {mol}^{- 1}) (298 K) \log K_{eq} or, \log K_{eq} = - \frac{107 \times 10^{3}}{2.303 \times 8.314 \times 298} or, \log K_{eq} = - 0.01875 \times 10^{3} or, K_{eq} = Antilog (- 18.75) = 1.77 \times 10^{- 19}

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