At a certain temperature, Kc = 33 for the reaction:
H2(g) + I2(g) 2HI(g)
Assume that the initial concentrations of both H2 and I2 are 6.00 x 10-3 mol/L. Find the concentration of each reactant and product at equilibrium.
You need to insert the arrow to know the difference between products and reactants. You have the equation, set up an ICE chart.
....................H2 + I2 ==> 2HI
...initial........0.006..0.006...0
change..............-x....-x.....2x
equilibrium.....0.006-x...0.006-x...2x
Now substitute into the Kc expression of
Kc = 0.33 = (HI)^2/(H2)(I2)
Solve for x and calculate each of the final concns. Post your work if you get stuck.
X=1.97406×10 power of -3 .then what is the reactant and product of equilibrium .can't understand actually .
To find the concentration of each reactant and product at equilibrium, we can make use of the equilibrium expression and the given value of Kc.
The equilibrium expression is given by:
Kc = [HI]^2 / ([H2] * [I2])
Given that Kc = 33, we can write the equation as:
33 = [HI]^2 / ([H2] * [I2])
Since the initial concentrations of H2 and I2 are both 6.00 x 10^-3 mol/L, we can substitute these values into the equation:
33 = [HI]^2 / (6.00 x 10^-3 * 6.00 x 10^-3)
Simplifying the equation, we have:
33 = [HI]^2 / 3.6 x 10^-5
Multiplying both sides of the equation by 3.6 x 10^-5, we get:
1.188 x 10^-3 = [HI]^2
Taking the square root of both sides, we find the concentration of HI:
[HI] = √(1.188 x 10^-3)
[HI] ≈ 0.0345 mol/L
Since the reaction is 1:1 stoichiometrically, the concentrations of H2 and I2 at equilibrium will be equal to each other. Therefore, the concentration of each reactant will be approximately:
[H2] ≈ 0.0345 mol/L
[I2] ≈ 0.0345 mol/L
To summarize, at equilibrium, the concentration of H2, I2, and HI will be approximately 0.0345 mol/L for each.
To find the concentration of each reactant and product at equilibrium, we need to use the given value of Kc and the initial concentrations.
Step 1: Write the balanced chemical equation and set up the ICE table.
The balanced chemical equation is: H2(g) + I2(g) ⇌ 2HI(g)
The ICE table represents the Initial, Change, and Equilibrium concentrations of each species involved in the reaction.
H2(g) + I2(g) ⇌ 2HI(g)
Initial: 6.00 x 10^-3 6.00 x 10^-3 0
Change: -x -x +2x
Equilibrium: 6.00 x 10^-3 - x 6.00 x 10^-3 - x 2x
Step 2: Set up the expression for Kc using the equilibrium concentrations.
The expression for Kc is given by the ratio of the products to the reactants, each raised to their respective stoichiometric coefficients.
Kc = ([HI]^2) / ([H2] * [I2])
Given that Kc = 33, we now have:
33 = (2x)^2 / ((6.00 x 10^-3 - x) * (6.00 x 10^-3 - x)).
Step 3: Solve for x.
To solve this equation, we rearrange it to a quadratic equation:
33(6.00 x 10^-3 - x)^2 - 4x^2 = 0.
Simplifying and solving the quadratic equation gives two possible solutions for x: x = 0.00133 M and x = 0.003 zM. However, we can discard the x = 0.003 zM solution since it exceeds the initial concentration, which is not physically possible. So, we consider x = 0.00133 M as the correct value for x.
Step 4: Calculate the equilibrium concentrations.
To find the equilibrium concentrations, substitute the value of x back into the ICE table equations.
[H2] at equilibrium = 6.00 x 10^-3 - x = 6.00 x 10^-3 - 0.00133 M = 0.00467 M
[I2] at equilibrium = 6.00 x 10^-3 - x = 6.00 x 10^-3 - 0.00133 M = 0.00467 M
[HI] at equilibrium = 2x = 2(0.00133 M) = 0.00266 M
Therefore, the concentration of each reactant and product at equilibrium is approximately:
[H2] = 0.00467 M
[I2] = 0.00467 M
[HI] = 0.00266 M