Consider the balanced equation 2A(g) + 3G(g) „\ 2X(g) + Z(g). To a 20.0 L container maintained at a temperature of 127 C were added 0.200 mol of A, 0.500 mol of G, 0.400 mol of X, and 0.600 mol of Z. After equilibrium was established, it was found that 0.340 mol of X was present. Calculate K for the reaction at 127 C.
I assume the "funny" symbol between G and 2X is an arrow. So it should look like this.
2A(g) + 3G(g) ==> 2X(g) + Z(g)
mols X changed from 0.400 mol before equilibrium was established to 0.34 at equilibrium. Thus X changed by 0.400-0.340 = 0.060 mols. That means that 0.060 mol X was used (and obviously the point of equilibrium is to the left). Now watch carefully, I will do this for one and you follow through with the others. If 0.060 mol was used from X, then 1/2 that number or 0.030 mol Z was used making the final mols Z = 0.600 - 0.030 = 0.570 mol Z at equilibrium. Then you know that A must have increased by 0.030 and G must have increased by 0.030 x (3mols G/2 mols X) = ??
Change all the mols to molarity (M = mols/20 L)and substitute into K = (X)^2(Z)/(A)^2(G)^3 = ?? to calculate Kc.
To calculate K for the reaction at 127°C, we need to use the given information and the equilibrium concentrations of the reactants and products.
First, let's write the expression for the equilibrium constant K:
K = [X(g)]^2 [Z(g)] / [A(g)]^2 [G(g)]^3
Given:
Initial moles of A = 0.200 mol
Initial moles of G = 0.500 mol
Initial moles of X = 0.400 mol
Initial moles of Z = 0.600 mol
Moles of X at equilibrium (after reaction) = 0.340 mol
To calculate the equilibrium concentrations, we need to consider the change in moles or concentrations of the reactants and products.
According to the balanced equation:
2A(g) + 3G(g) -> 2X(g) + Z(g)
From the stoichiometry, we can see that for every 2 moles of A used up, 2 moles of X are produced.
Similarly, for every 3 moles of G used up, 2 moles of X are produced.
Therefore, the change in moles of X is given by:
Change in moles of X = (initial moles of X) - (2 * moles of A used) - (2/3 * moles of G used)
Since we know the moles of X at equilibrium (0.340 mol), we can set up the equation:
0.340 = 0.400 - 2 * (0.200) - (2/3) * (0.500)
Now we can solve for the moles of G used:
0.340 = 0.400 - 0.400 - 1/3 * 0.500
0.340 = 0.400 - 0.133
0.340 = 0.267
Next, we can calculate the equilibrium concentrations by dividing the moles by the total volume of the container:
[A(g)] = 0.200 mol / 20.0 L = 0.010 M
[G(g)] = 0.500 mol / 20.0 L = 0.025 M
[X(g)] = 0.340 mol / 20.0 L = 0.017 M
[Z(g)] = 0.600 mol / 20.0 L = 0.030 M
Now we have all the necessary concentrations to calculate K:
K = [0.017]^2 * [0.030] / [0.010]^2 * [0.025]^3
K = 0.000289 * 0.030 / 0.0001 * 0.015625
K = 0.00000867 / 0.000001953125
K = 4.44
Therefore, the equilibrium constant (K) for the reaction at 127°C is approximately 4.44.