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B) a $50,0dm^{3}$ reaction vessel contains 1,00mol $n_{2}$ 0,500mol $nh_{3}$ and 3 $h_{2}$ Deduce what will happen to $nh_{3}$ when the mixture goes to equilibrium at $400^{circ }c$ The equation is $n_{2}(g)+2h_{2}(g)rightleftharpoons 2nh_{3(g)}$ $k=0,500$ at $400^{circ }c$ (6 marks)
Accepted Answer
To determine the change in ammonia concentration at equilibrium, we can use the reaction quotient (Q) and compare it to the equilibrium constant (K).
1. Calculate the initial reaction quotient (Q):
Q = ([NH3]^2) / ([N2]
[H2]^2)
Q = (0.500^2) / (1.00
3^2) = 0.0278
2. Compare Q to K:
Since Q (0.0278) is less than K (0.500), the system is not at equilibrium. The reaction will shift to the right (towards product formation) to reach equilibrium.
3. Equilibrium shift:
As the reaction shifts to the right, more ammonia (NH3) will be formed. The concentration of ammonia will increase.
Therefore, when the mixture reaches equilibrium at 400°C, the concentration of NH3 will increase.
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