4. N2+3H2----2Nh3 k=1.2 at a T=375C
you initially have 0.249mol of N2. 3.21*10^-2mol of H2, and 6.42*10^-4 mol of NH3 in a 3.5L container
a.what is the equation to determine the reaction quotient, Q
b.determine the reaction quotient Q
c. compare Q and K. determine which way the equilibrium will shift in the case. write out the new reaction demonstrating which way the shift has occurred
d.if the amount of Nh3 was increased to 6.42*10^-3mol, would the reaction shift?explain
e. if the amount of N2 was decreased to .200mol, would the reaction shift?explain.
The equation for determining the reaction quotient is the same as the equation for Keq. The only difference is that for Krxnquotient the values in the problem are substituted for concns in the Keq expression. The values that are substituted for Keq are the equilibrium values.
a. The equation to determine the reaction quotient, Q, is the same as the balanced chemical equation for the reaction. In this case, the balanced reaction equation is:
N2 + 3H2 -> 2NH3
So the equation for Q would be:
Q = [NH3]^2 / ([N2] * [H2]^3)
b. To determine the reaction quotient, Q, you need to plug in the concentrations of NH3, N2, and H2 into the reaction quotient equation. Given the concentrations in the question:
[NH3] = 6.42 * 10^-4 mol / 3.5 L
[N2] = 0.249 mol / 3.5 L
[H2] = 3.21 * 10^-2 mol / 3.5 L
Substituting the values into the equation:
Q = (6.42 * 10^-4 / 3.5)^2 / [(0.249 / 3.5) * (3.21 * 10^-2 / 3.5)^3]
c. To compare Q and K and determine which way the equilibrium will shift, we need to compare their values. Given that the equilibrium constant, K, is 1.2, we compare Q and K:
If Q > K, the reaction will shift to the left to reach equilibrium and decrease the concentration of products.
If Q < K, the reaction will shift to the right to reach equilibrium and increase the concentration of products.
If Q = K, the reaction is already at equilibrium.
d. If the amount of NH3 was increased to 6.42 * 10^-3 mol, the reaction will shift to the left to reach equilibrium. This is because the concentration of NH3 in the numerator of the reaction quotient Q will increase, while the concentrations of N2 and H2 in the denominator of Q remain constant. The shift to the left will decrease the value of Q, moving it closer to the equilibrium constant K.
The new reaction demonstrating the shift to the left would be:
2NH3 -> N2 + 3H2
e. If the amount of N2 was decreased to 0.200 mol, the reaction will shift to the right to reach equilibrium. This is because the concentration of N2 in the denominator of the reaction quotient Q will decrease, while the concentrations of NH3 and H2 in the numerator and denominator respectively remain constant. The shift to the right will increase the value of Q, moving it closer to the equilibrium constant K.
The new reaction demonstrating the shift to the right would still be:
N2 + 3H2 -> 2NH3