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  • Ammonia under a pressure of 15 atm, at 27oC is heated to 327oC in a closed vessel in the presence of catalyst. Under these conditions, NH3 partially

Ammonia under a pressure of 15 atm, at 27oC is heated to 327oC in a closed vessel in the presence of catalyst. Under these conditions, NH3 partially decomposes to H2 and N2.The vessel is such that the volume remains effectively constant, whereas the pressure increases to 50 atm. Calculate the percentage of NH3 actually decomposed.

Option: 1

50%


Option: 2

33.33%


Option: 3

66.7%


Option: 4

89%


Answers (1)

Given that the temperature is doubled, this means that if NH3 was not dissociated then the pressure would have been 30 atm. But after dissociation, the total pressure is 50 atm.

\text { Let } \alpha \text { be the degree of dissociation }


 Moles             2 \mathrm{NH}_{3}(\mathrm{g}) \rightleftharpoons 3 \mathrm{H}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g})
Initial                   a                         0              0
at equilibrium     a-a \alpha      3 a \alpha / 2    a \alpha / 2

\text { Total moles }=a+a \alpha

\frac{\text { Initial moles }}{\text { Final moles }}=\frac{\mathrm {Pressure\ of\ NH_3\ at\ 600 K }}{\text { Final pressure after dissociation at 600 K }}

\\\Rightarrow \frac{a}{a+a \alpha}=\frac{30}{50}\\\\ \Rightarrow \alpha=\frac{20}{30}

\\\% \text { dissociation }= \alpha \\\\\ \% \text { dissociation }=\frac{20}{30} \times 100=66.7 \%
Therefore,option(3) is correct

Posted by

Kshitij

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