For a certain reaction X $\stackrel{ }{\to }$ Y the value of $∆\mathrm{H} \mathrm{and} ∆\mathrm{S} \mathrm{are} 50.50 \mathrm{kJ} {\mathrm{mol}}^{-1} \mathrm{and} 100.03 {\mathrm{JK}}^{-1}$ respectively. The temperature t which $∆\mathrm{G}=0 \mathrm{is}$

Given, $∆\mathrm{H} = 50.50 \mathrm{kJ} {\mathrm{mol}}^{-1}$

$$\begin{gathered} ∆\mathrm{S} = 100.03 {\mathrm{JK}}^{-1}= 100\times {10}^{-3}=0.1 \mathrm{kJ} {\mathrm{K}}^{-1} \\ \therefore \mathrm{From} ∆\mathrm{G}=∆\mathrm{H}-\mathrm{T}∆\mathrm{S} \\ \Rightarrow 0=∆\mathrm{H}-\mathrm{T}∆\mathrm{S} \\ \Rightarrow \mathrm{T}=\frac{∆\mathrm{H}}{∆\mathrm{S}}=\frac{50.50}{0.1}=505 \mathrm{K} \\ \therefore \mathrm{T} = 505-273={232}^0\mathrm{C} \end{gathered}$$