Question

Given the ∆G° changes below, determine ∆G° for 2 A + B + D → 2 F

A + B → C ∆G° = -35 kJ

A + D → E + F ∆G° = 20 kJ

F → C + E ∆G° = 15 kJ

Answer 1

Solution :-

Given reaction equations

A + B → C ∆G° = -35 kJ

A + D → E + F ∆G° = 20 kJ

F → C + E ∆G° = 15 kJ

Desired reaction equation

2 A + B + D → 2 F

To get the desired reaction equation we need to rearrange the given three equations in such a way that after adding those three equations by cancelling the similar species from opposite side of the equations we get the desired equation.

Therefore we need to keep the equation 1 and 2 as it is and reverse the equation 3

When we reverse the equation then the value of the delta G remains same but the sign changes means positive becomes negative and negative changes to positive

Therefore new equations are as follows

A + B → C ∆G° = -35 kJ

A + D → E + F ∆G° = 20 kJ

C + E → F ∆G° = -15 kJ

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2A + B+D ---- > 2F               ∆G° = (-35 kJ + 20 kJ +(-15 kJ)) = -30 kJ

Therefore the delta g for the desired equation is -30 kJ