Chemical EquilibriumHard
Question
An amount of 3 moles of N2 and some H2 is introduced into an evacuated vessel. The reaction starts at t = 0 and equilibrium is attained at t = t1. The amount of ammonia at t = 2t1 is found to be 34 g. It is observed that $\frac{w\left( N_{2} \right)}{w\left( H_{2} \right)} = \frac{14}{3}\text{ at }t = \frac{t_{1}}{3}$and $t = \frac{t_{1}}{2}$. The only correct statement is
Options
A.w(N2) + w(H2) + w(NH3) = 118 g at t = t1
B.w(N2) + w(H2) + w(NH3) = 102 g at t = 2t1
C.w(N2) + w(H2) + w(NH3) = 50 g at t = t1/3
D.w(N2) + w(H2) + w(NH3) cannot be predicted
Solution
If reactants are taken in stoichiometric amount, then their mass ratio does not change at any stage of reaction. For 3 mole N2, there should be 9 mole H2. Hence, at any stage,
$m_{N_{2}} + m_{H_{2}} + m_{NH_{3}} = 3 \times 28 + 9 \times 2 = 102\text{ gm}$
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