Data given for the following reaction is as follows: $$\mathrm{FeO}_{(\mathrm{s})}+\mathrm{C}_{(\text {graphite })} \longrightarrow \mathrm{Fe}_{(\mathrm{s})}+\mathrm{CO}_{(\mathrm{g})}$$ | Substance | $\Delta_{\mathrm{f}} \mathrm{H}^{0}\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ | $\Delta \mathrm{S}^{0}\left(\mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$ | |-----------|------------|------------| | $\mathrm{FeO}_{(\mathrm{s})}$ | -266.3 | 57.49 | | $\mathrm{C}_{\text {(graphite) }}$ | 0 | 5.74 | | $\mathrm{Fe}_{(\mathrm{s})}$ | 0 | 27.28 | | $\mathrm{CO}_{(\mathrm{g})}$ | -110.5 | 197.6 | The minimum temperature in K at which the reaction becomes spontaneous is ____. (Integer answer)

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Chemical Thermodynamics

Data given for the following reaction is as follows:

FeO_{(s)} + C_{(graphite)} ⟶ Fe_{(s)} + CO_{(g)}

The minimum temperature in K at which the reaction becomes spontaneous is ____. (Integer answer)

Answer: 964

The reaction becomes spontaneous when

Δ G = Δ H - T Δ S

turns negative, so the minimum temperature is where

Δ G = 0

, i.e.

T = Δ H /Δ S

Δ H = [0 + (- 110.5)] - [(- 266.3) + 0] = 155.8 kJ mo l^{- 1},

Δ S = [27.28 + 197.6] - [57.49 + 5.74] = 161.65 J mo l^{- 1} K^{- 1} .

Then

T = \frac{155800}{161.65} = 964 K .

The reaction becomes spontaneous at

964 K

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer