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If the pressure of $N_2/H_2$ mixture in a closed vessel is 100 atm and 20% of the mixture then reacts , the pressure at the same temperature would be

$(a)\;The\; same\qquad(b)\;110\;atm\qquad(c)\;90\;atm\qquad(d)\;80\;atm$

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A)
Toolbox:
  • The ideal gas equation is : $\quad\large pV=nRT$
The reaction shown is:
$N_2+3H_2 = 2NH_3$
With a deduction of 4 moles of the $N_2/H_2$ mixture, 2 moles of Ammonia is obtained.
So, from 20% of the $N_2/H_2$ mixture, 10% Ammonia is produced, reducing the total mixture by 10%.
Given, the system is a closed vessel, Volume, V = constant and temperature, T = 100 atm.
$\therefore \quad \large p \propto n$
Since, the total mixture left is 90%, there will be a 10% decrease in the pressure on the mixture i.e. 90 atm.
Hence answer is (c)
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