In a reaction P + Q → R + S, there is no change in entropy. Enthalpy change for the reaction (Δ H) is 12 kJ mol-1. Under what conditions, reaction will have negative value of free energy change?

Question

In a reaction P + Q → R + S, there is no change in entropy. Enthalpy change for the reaction (Δ H) is 12 kJ mol-1. Under what conditions, reaction will have negative value of free energy change? (A) If Δ H is positive (B) If Δ H is negative (C) If Δ H is 24 kJ mol-1 (D) If temperature of reaction is high

Tardigrade Admin · Accepted Answer

Δ G = Δ H-TΔ S, Δ S = 0 or TΔ S = 0 Δ G = Δ H

Q. In a reaction $P + Q \to R + S$ , there is no change in entropy. Enthalpy change for the reaction $(Δ H)$ is $12 k J m o l^{- 1}$ . Under what conditions, reaction will have negative value of free energy change?

If $Δ H$ is positive

If $Δ H$ is negative

If $Δ H$ is $24 k J m o l^{- 1}$

If temperature of reaction is high

$Δ G = Δ H - T Δ S, Δ S = 0$ or $T Δ S = 0$
$Δ G = Δ H$

Q. In a reaction P+Q→R+S, there is no change in entropy. Enthalpy change for the reaction (ΔH) is 12kJmol−1. Under what conditions, reaction will have negative value of free energy change?

If ΔH is positive

If ΔH is negative

If ΔH is 24kJmol−1