Question

For a particular reaction, ΔH = -30.9 kJ and ΔS = -120.8 JK−1. Assume that ΔH and ΔS do not vary with temperature. At what temperature (in Kelvin) will the reaction have ΔG = 0?

Answer 1

We know the equation which relates `deltaG,deltaH,deltaS`

i.e `deltaG =deltaH -T deltaS`

So,putting, `deltaG=0`

we get, `T=(deltaH)/(deltaS)=(-30.9*10^3)/(-120.8)=255.8K`