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Question: The free energy change for the following reactions are given below \(C_{2}H_{2}(g) + \frac{5}{2}O_{...

The free energy change for the following reactions are given below

C2H2(g)+52O2(g)→2CO2(g)+H2O(l)C_{2}H_{2}(g) + \frac{5}{2}O_{2}(g) \rightarrow 2CO_{2}(g) + H_{2}O(l); ΔGo=−1234kJ\Delta G^{o} = - 1234kJ

C(s)+O2(g)→CO2(g);ΔGo=−394kJC(s) + O_{2}(g) \rightarrow CO_{2}(g);\Delta G^{o} = - 394kJ

H2(g)+12O2(g)→H2OH_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O (l); ΔGo=−237kJ\Delta G^{o} = - 237kJ

What is the standard free energy change for the reactionH2(g)+2C(s)→C2H2(g)H_{2}(g) + 2C(s) \rightarrow C_{2}H_{2}(g)

A

– 209 kJ

B

– 2259 kJ

C

+2259 kJ

D

209 kJ

Answer

– 209 kJ

Explanation

Solution

ΔG=ΔG0(Product)−ΔG0(Reactants)=(−1234)−[−237+2(−394)]=−1234+1025=−209kJ\Delta G = \Delta{G^{0}}_{(\text{Product})} - \Delta{G^{0}}_{(\text{Reactants})} = ( - 1234) - \lbrack - 237 + 2( - 394)\rbrack = - 1234 + 1025\mathbf{=}\mathbf{-}\mathbf{209kJ}