Question

When you create the equation of a line by substituting $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ Δ G ∘ = Δ H ∘ − T Δ S ∘ into the equation $\Delta G^\circ=-RT\:\ln\left(K_c\right)$ Δ G ∘ = − R T ln ⁡ ( K c ) , and solve for ln (K_c), what are your y and x values?

Choices are: ΔH/ΔS, -R/T, 1/T, T, ln(K), -ΔH/R

Answer 1

Answer:

We can equate the above given equation as;

ΔG = ΔH -TΔS = - RTln(K_c)

or ln(K_c) = -ΔH/RT + ΔS/R  

Here, y = ln(K_c) and 1/T = x

Where, m = -ΔH/R and C = ΔS/R

Hence, y = mx + C or ln(Kc) = (-ΔH/R)1/T + ΔS/R

Please let me know, if you have any doubt by commenting below the answer.

Thanks