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  • Determine which of the following reactions at constant pressure represent surrounding that do work on the system <img alt="4NH_{3}\; \left ( g \right )+7O_{2}\; \left ( g \right )\; \righ

Determine which of the following reactions at constant pressure represent surrounding that do work on the system

4NH_{3}\; \left ( g \right )+7O_{2}\; \left ( g \right )\; \rightarrow 4NO_{2}\; \left ( g \right )+6H_{2}O\; \left ( g \right )

CO\; \left ( g \right )+2H_{2}\; \left ( g \right )\rightarrow CH_{3}OH\; \left ( l\right )

C\left ( s,\; graphite \right )+H_{2}O\; \left ( g \right )\; \rightarrow CO\; \left ( g \right )+H_{2}\; \left ( g \right )

H_{2}O\; \left ( s \right )\rightarrow H_{2}O\; \left ( l\right )

Option: 1

III, IV


Option: 2

II, III              


Option: 3

II, IV


Option: 4

I, II and IV


Answers (1)

best_answer

The reactions at constant pressure represent surroundings that do work on the system environment.

Work done on the system, so, work is positive.

Formula.

W=-P_{ext}.\Delta V=-\Delta n_{g}RT   

\Delta n_{g}  is (-ve) for I and II

also  \Delta \textup{V=-ve for IV}

because the volume of Ice (Solid water ) is more than the water (liquid state).

Another Solution

When gas is evolved during a chemical reaction, the gas can be imagined as displacing the atmosphere - pushing it back against the atmospheric pressure. The work done is therefore VXP where V is the volume of gas evolved, and P is the atmospheric pressure.

 

4 \mathrm{NH}_{3}(\mathrm{~g})+7 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})

In this reaction, 11 vols of reagent combine to form 10 vol of product. The total volume falls, so the work done is negative (the atmosphere does work on the gaseous system). 

 

\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})

3 volumes of reagent combine to form 1 vol of product.  The total volume falls, so the work done is negative (the atmosphere does work on the gaseous system).

 

\mathrm{C}(\mathrm{s}, \text { graphite })+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})

Here, 2 vols of reagents combine to produce 2 vols of product. No change in surroundings. 

 

\mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})

In this reaction, solid volumes of reagent combine to form the liquid product. The total volume increases, so the work done on the atmosphere is negative.

 

Therefore, Option(4) is correct.

Posted by

Pankaj

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