SATHEE: Equilibrium Question 624

Equilibrium Question 624

Question: 28 g of $N_{2}$ and 6 g of $H_{2}$ were kept at $400^{o} C$ in 1 litre vessel, the equilibrium mixture contained $27.54 g$ of $N H_{3}$ . The approximate value of $K_{c}$ for the above reaction can be (in $m o l e^{- 2} l i t r e^{2}$ ) [CBSE PMT 1990]

Options:

A) 75

B) 50

C) 25

D) 100

Show Answer

Answer:

Correct Answer: A

Solution:

$N_{2} + 3 H_{2}$ ⇌ $2 N H_{3}$ Initial conc. 1 3 0 at equilibrium 1-0.81 3-2.43 1.62 0.19 0.57 No. of moles of $N_{2} = \frac{28}{28} = 1$ mole No. of moles of $H_{2} = \frac{6}{2} = 3$ mole No. of moles of $N H_{3} = \frac{27.54}{17} = 1.62$ mole $K_{c} = \frac{{[N H_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{[1.62]}^{2}}{[0.19] {[0.57]}^{3}}$ =75

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Equilibrium Question 624

Question: 28 g of N2 and 6 g of H2 were kept at 400oC in 1 litre vessel, the equilibrium mixture contained 27.54g of NH3 . The approximate value of Kc for the above reaction can be (in mole−2litre2 ) [CBSE PMT 1990]

Options:

Answer:

Solution:

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Question: 28 g of $N_{2}$ and 6 g of $H_{2}$ were kept at $400^{o} C$ in 1 litre vessel, the equilibrium mixture contained $27.54 g$ of $N H_{3}$ . The approximate value of $K_{c}$ for the above reaction can be (in $m o l e^{- 2} l i t r e^{2}$ ) [CBSE PMT 1990]