if the equilibrium concentration of nitrogen at 18degrees C is 0.256M and equil. concen. of hydrogen is 0.0184M what is the equil. concen. of ammonia?
N2 + 3 H2 <=> 2 NH3;
so K= [NH3]^2 / [N2][H2]^3
Is this correct?
Then ICE table? Giving me....
K= (2x)^2 / (0.256-x)(0.0814 -3x)^3
Am I on the right track?
You have substituted 0.256-x as the equilibrium concn of N2 but the problem states it is 0.256M. You have substituted 0.0184-3x as the equil concn of H2 but the problem states it is 0.0184M.
Oh and K=.0244
so is this correct?
.0244 = (2x)^2 / (0.256)(0.0814)^3
x= 9x10^-4
Yes, you are on the right track! The expression you wrote for the equilibrium constant (K) is correct:
K = [NH3]^2 / [N2][H2]^3
To solve for the equilibrium concentration of ammonia (NH3), you can use the ICE table approach. Here's how you can set up the ICE table:
Initial Concentrations:
[N2] = 0.256 M
[H2] = 0.0184 M
[NH3] = 0 (since none of it is present initially)
Change in Concentrations:
[N2] changes by -x (since it is being consumed in the reaction)
[H2] changes by -3x (since it is being consumed in the reaction)
[NH3] changes by +2x (since two moles of NH3 are formed for every mole of N2 consumed)
Equilibrium Concentrations:
[N2] = 0.256 - x
[H2] = 0.0184 - 3x
[NH3] = 2x
Now, you can substitute these concentrations into the equilibrium constant expression and solve for x.
K = (2x)^2 / ((0.256 - x)(0.0184 - 3x)^3
Now, you can solve this equation to find the value of x, and then use that value to calculate the equilibrium concentration of NH3 (2x).