The free energy change for a reversible reaction at equilibrium is:

zero

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positive

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negative

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none of these

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Solution

The correct option is A zero
At equilibrium, the value of the Gibb's free energy, also called as free energy change, is zero.
Hence, answer is option A.

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Similar questions

The free energy change of a reversible reaction at equilibrium is

log K = \frac{- Δ G^{0}}{2.303} RT

R = \frac{25}{3} J/molK, \frac{1}{12} L atm/molk, ln 2 = 0.7, ln 3 = 1.1, ln 5 = 1.6, log 2 = 0.3, log 3 = 0.5, l o g 5 = 0.7

N_{2} O_{4}

N O_{2}

27^{\circ}

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