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  • For the reaction: <img alt="\mathrm{2 \mathrm{NO}(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}_2(g)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2%20%5Cmathrm%7BNO%7

For the reaction:

\mathrm{2 \mathrm{NO}(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}_2(g)}

Given that the standard Gibbs free energy change\mathrm{\Delta G^{\circ}} is -150 kJ/mol at 25°C and the standard entropy change \mathrm{\Delta S^{\circ}}is 150 J/(mol·K), determine if the reaction is spontaneous at 25°C.

Option: 1

Spontaneous at 298 K
 


Option: 2

Spontaneous at below 298 K
 


Option: 3

Non spontaneous at 298 K
 


Option: 4

Spontaneous above 298 K


Answers (1)

best_answer

The spontaneity of a reaction can be determined using the relationship between
the standard Gibbs free energy change \mathrm{(\Delta G^{\circ})}, standard enthalpy change \mathrm{\Delta H^{\circ}}and standard entropy change \mathrm{\Delta S^{\circ}}by the equation:

\mathrm{\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}}

where T is the temperature in Kelvin.

Given

\mathrm{\begin{aligned} \Delta G^{\circ} & =-150 \mathrm{~kJ} / \mathrm{mol} \\ \Delta S^{\circ} & =150 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ T & =25^{\circ} \mathrm{C}=298 \mathrm{~K} \end{aligned}}

Substituting the values:

\mathrm{\begin{gathered} \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\ -150 \mathrm{~kJ} / \mathrm{mol}=\Delta H^{\circ}-298 \mathrm{~K} \times(0.150 \mathrm{~kJ} / \mathrm{mol}) \end{gathered}}

Solving for \mathrm{\Delta H^{\circ}}

\mathrm{\Delta H^{\circ}=-150 \mathrm{~kJ} / \mathrm{mol}+44.7 \mathrm{~kJ} / \mathrm{mol}=-105.3 \mathrm{~kJ} / \mathrm{mol}}

Since \mathrm{\Delta H^{\circ}} is negative and \mathrm{\Delta S^{\circ}} is positive, the reaction is spontaneous at25°C.

So, option A is correct

Posted by

Sanket Gandhi

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