At the start of the reaction there are 0.714 mole of H2, 0.984 mole of I2, and 0.886 mole of HI in a 2.70 L reaction chamber. Calculate the concentrations of the gases at equilibrium.
(H2) = 0.714/2.70 = ? about 0.26M
(I2) = 0.984/2.70 = ?about 0.36
(HI) = 0.886/2.70 = ?about 0.33
This is as far as you an go. You need an equilibrium constant.
(H2) = 0.714/2.70 = ? about 0.26M
(I2) = 0.984/2.70 = ?about 0.36
(HI) = 0.886/2.70 = ?about 0.33
You need to recalculate the above and be more accurate.
............H2 + I2 ==> 2HI
I........0.26M..0.36....0.33
C..........?.....?........?
First you must decide which way the reaction will go; i.e., left or right. To do this you calculate Qrxn.
Qrxn = (HI)^2/(H2)(I2)
Q = (0.33)^2/(0.26)(0.36) = about 1.16.
Compared with K = 53 that means the products are too small and reactants too large so the reaction will go to the right.
C...........-x.......-x......+x
E.........0.26-x...0.36-x ...0.33+2x
Substitute the E line into Keq expression and solv.
To calculate the concentrations of the gases at equilibrium, we first need to determine the balanced chemical equation for the reaction. Let's assume that the reaction is:
H2(g) + I2(g) ⇌ 2HI(g)
According to the balanced chemical equation, the stoichiometry of the reaction is 1 mole of H2 reacts with 1 mole of I2 to produce 2 moles of HI. From the given initial amounts of substances, we can construct an ICE (Initial, Change, Equilibrium) table to help solve the problem.
1. Initial: The initial amounts of the substances:
- H2: 0.714 mole
- I2: 0.984 mole
- HI: 0.886 mole
2. Change: The changes in the amount of substances (x) during the reaction:
- H2: -x (because 1 mole of H2 reacts to produce 2 moles of HI, so the change is -x)
- I2: -x (because 1 mole of I2 reacts to produce 2 moles of HI, so the change is -x)
- HI: +2x (because 1 mole of H2 and 1 mole of I2 react to produce 2 moles of HI, so the change is +2x)
3. Equilibrium: The amounts of substances at equilibrium:
- H2: 0.714 - x
- I2: 0.984 - x
- HI: 0.886 + 2x
Now, we need to use the provided information to find the value of x, which represents the extent of the reaction. We know that at equilibrium, the total volume of the reaction chamber is 2.70 L. Therefore, the sum of the equilibrium concentrations should be equal to the total volume of the chamber.
Equation: [H2] + [I2] + [HI] = total volume
Substituting the equilibrium concentrations:
(0.714 - x) + (0.984 - x) + (0.886 + 2x) = 2.70
Simplifying the equation:
2.584 + 2x = 2.70
Rearranging the equation:
2x = 2.70 - 2.584
2x = 0.116
x = 0.058
Now that we have the value of x, we can calculate the concentrations at equilibrium:
[H2] = 0.714 - x = 0.714 - 0.058 = 0.656 mole/L
[I2] = 0.984 - x = 0.984 - 0.058 = 0.926 mole/L
[HI] = 0.886 + 2x = 0.886 + 2(0.058) = 0.886 + 0.116 = 1.002 mole/L
Therefore, the concentrations of the gases at equilibrium are:
[H2] = 0.656 mole/L
[I2] = 0.926 mole/L
[HI] = 1.002 mole/L