given the reacting system: H2 (G) + I2(G) ---> 2HI(g) Keq= 64
Equal moles of H2, I2, and HI are placed in a 1.0 L container. Use calculations to determine the direction the reaction will proceed in order to reach equilibirum
..........H2 + I2 ==> 2HI
Keq = 64 = (HI)^2/(H2)((I2)\
Try any number, say 1
Qrxn = (1)^2/1*1 = 1
Try 2.
(2^2)/2*2 = 1
Try 3.
3^2/3*3 = 1
Get the picture?
So Q = 1; K = 64. That means (from Q) that the numerator is too small and the denominator is too large (that's the only way you can get a smaller number than 64). So H2 and I2 are too large; HI is too small. Which way MUST it go?
To determine the direction in which the reaction will proceed to reach equilibrium, we need to compare the initial concentrations of the reactants and products with their equilibrium concentrations. We can then use the equilibrium constant expression to calculate the reaction quotient (Q) and compare it to the equilibrium constant (Keq).
Given:
H2 (G) + I2 (G) ⇌ 2HI (G)
Keq = 64
Let's assume that initially, each species has a concentration of 'x' moles per liter.
The reaction quotient (Q) can be calculated by substituting the initial concentrations into the equilibrium constant expression:
Q = [HI]^2 / ([H2] * [I2])
Since the moles of H2, I2, and HI are equal initially, we can write:
Q = (x^2) / (x * x) = x^2 / x^2 = 1
Comparing the reaction quotient (Q) to the equilibrium constant (Keq), we have:
Q < Keq
Since Q is less than Keq, the reaction will proceed in the forward direction to reach equilibrium. This means that more HI will be formed until the concentrations of H2, I2, and HI are no longer changing, and the system reaches equilibrium.
To determine the direction the reaction will proceed to reach equilibrium, you can compare the initial concentrations of the reactants with the equilibrium constant (Keq) value. In this case, the equilibrium constant (Keq) is given as 64.
First, let's assign a variable to the initial concentration of H2, I2, and HI, let's call it "x" (since they are all equal).
Then, we can write the initial concentrations as follows:
[H2] = x
[I2] = x
[HI] = 2x (according to the stoichiometry of the balanced equation)
Now, let's calculate the value of Q, the reaction quotient, which is the ratio of the concentrations of the products to the reactants at any point in the reaction.
Q = ([HI]^2) / ([H2] * [I2])
Substituting the given initial concentrations, we have:
Q = ([2x]^2) / (x * x)
Q = 4x^2 / x^2
Q = 4
Comparing the value of Q (which is 4) with the value of Keq (which is 64), we can determine the direction the reaction will proceed:
- If Q < Keq (4 < 64), it means that there are more reactants than products, and the reaction will shift to the right (towards the formation of more products) to reach equilibrium.
- If Q > Keq (4 > 64), it means that there are more products than reactants, and the reaction will shift to the left (towards the formation of more reactants) to reach equilibrium.
In this case, since Q < Keq (4 < 64), the reaction will shift to the right (towards the formation of more products) to reach equilibrium.