Which of the following processes will never b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.The relationship is given by $\Delta G =

Vedclass · Accepted Answer

$2Ag + 3N_2 ightarrow 2AgN_3, \Delta H = 	ext{positive}, \Delta S = 	ext{negative}$

Vedclass · Answer

(D) For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G The relationship is given by $\Delta G = \Delta H - T\Delta S$.For option $D$,$\Delta H$ is positive and $\Delta S$ is negative.Therefore,$\Delta G = (+\Delta H) - T(-\Delta S) = (+\Delta H) + (T\Delta S)$.Since both terms are positive,$\Delta G$ will always be positive at all temperatures.Thus,the process will never be spontaneous.

Which of the following processes will never be a spontaneous process?

Similar Questions

For vaporization of water at $1$ atmospheric pressure,the values of $\Delta H$ and $\Delta S$ are $40.63 \ kJ \ mol^{-1}$ and $108.8 \ J \ K^{-1} \ mol^{-1},$ respectively. The temperature when Gibb's energy change $(\Delta G)$ for this transformation will be zero,is ............ $K$. (in $.4$)

In which of the following conditions can a chemical reaction not occur?

If a process is non-spontaneous,which of the following options is correct?

Calculate $\Delta H^{\circ}$ for a reaction at $298 \ K$ if $\Delta S^{\circ} = 120 \ J \ K^{-1}$ and $\Delta G^{\circ} = 28000 \ J$. (in $kJ$)

Calculate the value of $\Delta G$ for the following reaction: $N_2O_{4(g)} \longrightarrow 2NO_{2(g)}$ if $\Delta H = 57.44 \ kJ$ and $\Delta S = 176 \ J \ K^{-1} \ mol^{-1}$ at $300 \ K$. (in $kJ$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module