Liquid Nitrogen tetroxide, N2O4(l), was used as a fuel in Apollo missions to the moon. In a closed container the gas N2O4(g) decomposes to nitrogen dioxide, NO2(g). The equilibrium constant,k, for this reaction is 0.87 at 55 degrees Celsius. A vessel filled with N2O4(g) at this temperature is analyzed twice during the course of the reaction and found to contain the following concentrations:
(a) [N2O4(g)] = 5.30 mol/L; [NO2(g)] = 2.15 mol/L
(b) [N2O4(g)] = 0.80 mol/L;[NO2(g)]] = 1.55 mol/L
In each case determine wheter the system. If not, predict the direction in which the reaction will proceed to acheive equilibrium.
You didn't finish your question sentence. I assume you meant to write, "In each case determine whether the system 'is in equilibrium.'". I fnot......
determine the equilibrium quotient for each case.
(a) Kquot = (NO2)^2/(N2O4)= (2.15)^2/(5.30) = 0.872 which is essentially the same as 0.87 so the system is in equilibrium. I suppose, if we wish to get picky, that 0.872 is SLIGHTLY larger than Keq, so the system will shift ever so slightly to the left.
(b) Kquot = 1.55^2/0.80 = 3.00. Therefore, the system is not in equilibrium. We get a numbre larger than Keq because the products are too large and the reactants too small. That means the reaction must shift to the left to achieve equilibrium.
To determine whether the system is at equilibrium or not, we need to compare the given concentrations with the equilibrium concentrations.
First, let's write the balanced chemical equation for the reaction:
N2O4(g) ⇌ 2NO2(g)
The equilibrium expression is given as:
K = [NO2(g)]^2 / [N2O4(g)]
Let's calculate the values of the equilibrium concentrations using the given information:
(a) [N2O4(g)] = 5.30 mol/L; [NO2(g)] = 2.15 mol/L
Using the equilibrium expression, we find:
K = (2.15 mol/L)^2 / (5.30 mol/L) ≈ 0.872
Since the calculated value of K is approximately equal to the given value of 0.87, the system is at equilibrium.
(b) [N2O4(g)] = 0.80 mol/L; [NO2(g)] = 1.55 mol/L
Calculating the value of K using the equilibrium expression:
K = (1.55 mol/L)^2 / (0.80 mol/L) ≈ 3.02
Since the calculated value of K is not equal to the given value of 0.87, the system is not at equilibrium.
To predict the direction in which the reaction will proceed to achieve equilibrium, we can compare the calculated value of K to the given value of K. If K > 0.87, the reaction will shift towards the reactants (left); if K < 0.87, the reaction will shift towards the products (right).
In case (b), since K = 3.02 > 0.87, the reaction will proceed in the reverse direction to reduce the concentration of products (NO2) and increase the concentration of reactants (N2O4) to reach equilibrium. Hence, the reaction will shift to the left.
To determine whether the system is at equilibrium or not, we need to compare the given concentrations with the equilibrium constant.
The balanced equation for the decomposition of N2O4 is:
N2O4(g) ⇌ 2NO2(g)
The equilibrium constant expression for this reaction is:
K = [NO2(g)]^2 / [N2O4(g)]
Let's analyze the given concentrations and calculate the Q (reaction quotient) for both cases.
(a) For the first case, [N2O4(g)] = 5.30 mol/L and [NO2(g)] = 2.15 mol/L:
Q = ([NO2(g)]^2) / [N2O4(g)] = (2.15^2) / 5.30 ≈ 0.875
Comparing Q (0.875) with the equilibrium constant (K = 0.87), we see that they are very close. Since Q is not significantly different from K, we can conclude that the system is at equilibrium for the first case.
(b) For the second case, [N2O4(g)] = 0.80 mol/L and [NO2(g)] = 1.55 mol/L:
Q = ([NO2(g)]^2) / [N2O4(g)] = (1.55^2) / 0.80 ≈ 3.01
Comparing Q (3.01) with the equilibrium constant (K = 0.87), we see that they are significantly different. The system is not at equilibrium for the second case.
To predict the direction in which the reaction will proceed to achieve equilibrium, we compare the Q value with the equilibrium constant (K).
If Q < K, it means that there are more reactants (N2O4) than products (NO2), so the forward reaction is favored. The reaction will proceed in the forward direction to form more products until Q reaches the value of K.
If Q > K, it means that there are more products (NO2) than reactants (N2O4), so the reverse reaction is favored. The reaction will proceed in the reverse direction to form more reactants until Q reaches the value of K.
In the second case, we have Q = 3.01, which is greater than K. Therefore, the reaction will proceed in the reverse direction (from NO2 to N2O4) to form more reactants until the system reaches equilibrium.