Calculate the equilibrium concentrations of N2, H2, and NH3 present when a mixture that was initially 0.10M N2, 0.10 M H2, and 0.10 M NH3 comes to equilibrium at 500 degrees C.
N2(g) + 3H2(g) <--> 2NH3(g)
Kc = 0.040 (at 500 degrees C)
I'm getting confused with the math.
Why don't you show the math so we will know what is confusing you instead of us guessing at the problem. Here is the chemistry.
.........N2 + 3H2 ==> 2NH3
I.......0.1...0.1......0.1
C.......-x....-3x......2x
E......0.1-x..0.1-3x...2x
Kc = 0.04 = (NH3)^2/(N2)(H2)^3
0.040 = (.1+2x)^2 / (.1-x)(.1-3x)^3
-.00996-.4012x-3.9712x^2+1.08x^3
I'm not sure how to solve for x. I got the chemistry correct
the second line equals 0
To calculate the equilibrium concentrations of N2, H2, and NH3, you need to use the equilibrium constant expression and the initial concentrations of the reactants. Here's how you can do it step by step:
1. Write down the balanced equation for the reaction:
N2(g) + 3H2(g) ↔ 2NH3(g)
2. Write down the equilibrium constant expression(Kc) for the reaction:
Kc = [NH3]^2 / ([N2] * [H2]^3)
3. Substitute the initial concentrations into the expression:
Kc = [NH3]^2 / ([N2] * [H2]^3)
Kc = (x)^2 / ((0.10 M - x) * (0.10 M - 3x)^3)
Note: Since the reaction has coefficients of 1, 3, and 2, you assume that the initial concentration of N2 decreases by x, the initial concentration of H2 decreases by 3x, and the initial concentration of NH3 increases by 2x at equilibrium.
4. Use the given equilibrium constant(Kc) to solve for x:
Kc = 0.040
0.040 = (x)^2 / ((0.10 M - x) * (0.10 M - 3x)^3)
Solve this equation to find the value of x. You can use numerical methods, such as trial and error or a solver, to find the approximate value of x.
5. Once you have the value of x, you can calculate the equilibrium concentrations:
[N2] = 0.10 M - x
[H2] = 0.10 M - 3x
[NH3] = 0.10 M + 2x
Substitute the value of x into these expressions to calculate the equilibrium concentrations.
Note: The value of x represents the change in concentration from the initial concentrations to the equilibrium concentrations.