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  • Determine the standard entropy change for the reaction: <img alt="\mathrm{ \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3(g) }" src="https://entrancecorner.oncodecogs.com/gif.

Determine the standard entropy change for the reaction:
\mathrm{ \mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3(g) }
at \mathrm{ 300 \mathrm{~K} }. Given the standard entropies of \mathrm{ \mathrm{N}_2(g), \mathrm{H}_2(g), \: and \: \mathrm{NH}_3(g) } are \mathrm{ 191.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}, 130.6 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} }, and \mathrm{ 192.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}, } respectively.
 

Option: 1

200 \mathrm{~J} / \mathrm{K}-\mathrm{mol}
 


Option: 2

198.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol}
 


Option: 3

176


Option: 4

-


Answers (1)

Step 1: Calculate the standard entropy change \mathrm{\left(\Delta S^{\circ}\right)} for the reaction using the standard entropies of the reactants and products.

\mathrm{ \Delta S^{\circ}=\sum\left(n \cdot S_{\text {products }}^{\circ}\right)-\sum\left(n \cdot S_{\text {reactants }}^{\circ}\right) }

Step 3: Substitute the given standard entropies and calculate the total standard entropy change \mathrm{ \left(\Delta S_{\text {total }}^{\circ}\right). }

\mathrm{ \Delta S_{\text {total }}^{\circ}=[(2 \mathrm{~mol}) \cdot(192.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol})]-[(1 \mathrm{~mol}) \cdot(191.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol})+(3 \mathrm{~mol}) \cdot(130.6 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol})] }

Calculating the numerical value of the total standard entropy change:

\mathrm{ \Delta S_{\text {total }}^{\circ}=-198.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol} }

So, the standard entropy change for the reaction is \mathrm{ - 198.3 \mathrm{~J} / \mathrm{K}-mol }.

So, the correct option is 2.

Posted by

Ramraj Saini

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