Now,

\( \Delta G = \Delta H – T\Delta S\)

Let, the given reaction is at equilibrium, then \( \Delta T\) will be:

T =\( (\Delta H – \Delta G)\dfrac{1}{\Delta S}\dfrac{\Delta H}{\Delta S}\)

(\( \Delta G\) = 0 at equilibrium)

= \( \dfrac{ 400kJ mol^{-1}}{0.2kJ K^{-1}mol^{-1}}\)

Therefore, T = 2000K

Thus, for the spontaneous, \( \Delta G\) must be –ve and T > 2000K.

Answered by Pragya Singh | 1 year agoCalculate the entropy change in surroundings when 1.00 mol of \( H_2O\)(l) is formed under standard conditions. \( Δ_f H^ Θ= –286 kJ mol^{–1}.\)

Comment on the thermodynamic stability of NO(g), given,

(\( \dfrac{1}{2}\))N_{2(g)} + (\( \dfrac{1}{2}\))O_{2(g) }→ NO_{(g)}; \( \Delta _{r}H^{\Theta } = 90kJmol^{-1}\)

NO_{(g)} + (\( \dfrac{1}{2}\))O_{2(g)} → NO_{2(g)}; \( \Delta _{r}H^{\Theta } = -74kJmol^{-1}\)

The equilibrium constant for a reaction is 10. What will be the value of ∆G_{0}? R = 8.314 JK^{–1} mol^{–1}, T = 300 K.

For the reaction

2A_{(g)} + B_{(g)}→ 2D_{(g)}

\( \Delta U^{\Theta } = -10.5 kJ \;and \;\Delta S^{\Theta } = -44.1JK^{-1}\)

Calculate \( \Delta G^{\Theta }\) for the reaction, and predict whether the reaction may occur spontaneously.

For the reaction

2Cl_{(g)}→ Cl_{2(g)}

What are the signs of \( \Delta S\; and \;\Delta H\)?