At 1473K the equilibrium constant for the reaction H2(g) + Cl2(g) <--------> 2HCl(g)
Kc = 2.5*10^4. What would be the equilibrium concentration of HCL if the initial concentrations of H2 and Cl2 were 0.1840M and 0.1918M. I got 29.7 but was told it was wrong
..........H2 + Cl2 ==> 2HCl
begin..0.1840 0.1918...0
change.....-x...-x.......+2x
final.0.1840-x..0.1918-x...2x
Substitute into Kc and solve for x, then multiply by 2 for HCl.
To find the equilibrium concentration of HCl, we can use the given equilibrium constant and the initial concentrations of H2 and Cl2.
First, let's define the changes in concentration for each species at equilibrium:
[H2] = initial concentration of H2 - (2 * change in [HCl])
[Cl2]= initial concentration of Cl2 - (change in [HCl])
[HCl] = 2 * change in [HCl]
Now, let's assume the change in [HCl] = x. Since two moles of HCl are produced for every mole of H2 consumed, the change in [H2] and [Cl2] will be -2x and -x, respectively.
Next, substitute the values into the equilibrium expression:
Kc = [HCl]^2 / ([H2] * [Cl2])
2.5*10^4 = (2x)^2 / ((0.1840 - 2x) * (0.1918 - x))
Now, solve the equation for x.
(2.5*10^4) * (0.1840 - 2x) * (0.1918 - x) = (2x)^2
(2.5*10^4) * (0.1840 - 2x) * (0.1918 - x) = 4x^2
Rewrite the equation in standard form:
10^4 * (0.460 - 5.0x + 4x^2) = 4x^2
10^4 * 4x^2 - 10^4 * 5.0x + 10^4 * 0.460 = 4x^2
40x^2 - 5.0x*10^4 + 4.6*10^3 = 4x^2
36x^2 - 5.0x*10^4 + 4.6*10^3 = 0
Now, solve the quadratic equation. We can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values:
x = (-(-5.0x10^4) ± √((-5.0x10^4)^2 - 4(36)(4.6x10^3))) / (2*36)
x = (5.0x10^4 ± √(25x10^8 - 663.36x10^3)) / 72
x = (5.0x10^4 ± √(25x10^8 - 663.36x10^3)) / 72
x = (5.0x10^4 ± √(24.99x10^8)) / 72
x = (5.0x10^4 ± 4.999x10^4) / 72
Now, we have two possible values of x:
1) x = (5.0x10^4 + 4.999x10^4) / 72 = 6.944x10^4 / 72 = 965.6 M
2) x = (5.0x10^4 - 4.999x10^4) / 72 = 1 M
Since the equilibrium concentrations cannot be negative, we reject the second value.
Therefore, the equilibrium concentration of HCl is 965.6 M.
To find the equilibrium concentration of HCl, we need to use the given initial concentrations of H2 and Cl2, along with the equilibrium constant (Kc). The equation for the reaction tells us that 1 mole of H2 reacts with 1 mole of Cl2 to form 2 moles of HCl.
First, let's write the expression for the equilibrium constant (Kc):
Kc = [HCl]^2 / ([H2] x [Cl2])
Now, let's substitute the given values into the equation:
Kc = (x)^2 / (0.1840 * 0.1918)
Since the equilibrium concentration of HCl is not given, we can assume it to be 'x'.
Now, we need to solve for 'x'. Rearranging the equation, we get:
x^2 = Kc * (0.1840 * 0.1918)
x^2 = 2.5 * 10^4 * (0.1840 * 0.1918)
x^2 = 892.03
Taking the square root of both sides, we find:
x = √892.03
x ≈ 29.8 M
Therefore, the equilibrium concentration of HCl would be approximately 29.8 M, not 29.7 M as you calculated.