Thermodynamic+Chemistry

Chemistry Level 4

Consider the above cycle where the gas is diatomic.

Find the efficiency of the cycle ABCDA.( = η = \eta ) .

Now consider the following reaction(Unbalanced)

N X 2 + H X 2 N H X 3 \ce{N2 + H2 \rightarrow NH3}

If the percentage yeild of the above reaction is 3100 8 η \dfrac{3100}{8} \eta .

At t = 0 t=0 , 2 2 moles of N X 2 \ce{N2} and 8.9 8.9 moles of H X 2 \ce{H2} .

If N N moles of N H X 3 \ce{NH3} is obtained.

Then it is used in the following reaction to get product of [ A g ( N H X 3 ) X 2 ] X + \ce{[Ag(NH3)2]+} .

The following reaction(unbalanced) was used for preparation of [ A g ( N H X 3 ) X 2 ] X + \ce{[Ag(NH3)2]+} .

A g X + + N H X 3 [ A g ( N H X 3 ) X 2 ] X + \ce{Ag+ + NH3 \rightarrow [Ag(NH3)2]+} (Percentage yeild 100%) .

If the volume of container used in preparation of [ A g ( N H X 3 ) X 2 ] X + \ce{[Ag(NH3)2]+} is 2 L 2L .

Find concentration of [ A g ( N H X 3 ) X 2 ] X + \ce{[Ag(NH3)2]+} in molarity after end of reaction.


The answer is 0.5.

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1 solution

Ratio of isobaric and isochoric specific heats of a diatomic gas is (7/5). The first law efficiency of the given cycle is (4/31). Thus the reaction yield of the first reaction is 50%. Therefore 2 moles of ammonia is produced in the reaction, and 1 mole of the ion is produced in the container. Since the volume of the container is 2 litres, the required molarity is (1/2) or 0.5

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