What are the moles of substances present at equilibrium at 450C if 1 mol N2 and 4 mol H2 in a 10 L vessel react according to the following equation?
N2 (g) + 3H_2(g) ⇌ 2NH_3 (g) The equilibrium constant Kc is 0.153 at 450C.
Thanks a lot.
Kc=x^2/(.1-x/2)(.4-3x/2)
avoiding fractions, mutiply right hand side by 4/4
Kx=4x^2/(.2-x)(.8-6x)
.153(.16-1.2x-.8x+6x^2)=x^2
at this point, gather terms, put in quadratic form, and solve for x to get product concentration, and
.1-x/2 for concentration of N2
.4-3x/2 for concentration of H2
check my thinking.
I believe my friend made a couple of typos. I think the Kc expression should be 4x^2/(0.2-2x)(.8-6x)^3
where I have bolded the numbers I think it should be.
To find the moles of substances present at equilibrium, we need to use the information given about the initial moles and the equilibrium constant.
Given:
Initial moles of N2 = 1 mol
Initial moles of H2 = 4 mol
Equilibrium constant (Kc) = 0.153
The balanced chemical equation is:
N2 (g) + 3H2(g) ⇌ 2NH3(g)
From the equation, we can see that every 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.
At equilibrium, let's assume that x moles of N2 react with 3x moles of H2 to form 2x moles of NH3.
Using the given information, we can set up the equilibrium expression in terms of moles:
Kc = [NH3]^2 / ([N2] * [H2]^3)
Substituting the values into the expression:
0.153 = (2x)^2 / (1-x) * (4-3x)^3
Now, we need to solve this quadratic equation for x.
To do this, we can rearrange the expression and set it equal to zero:
0 = 0.153(4-3x)^3 - 4x^2
Next, we can expand and simplify the expression:
0 = 0.153(64-144x+108x^2-36x^3) - 4x^2
0 = 9.792 - 22.032x + 16.524x^2 -5.508x^3 - 4x^2
Combine like terms:
0 = -5.508x^3 + 12.524x^2 - 22.032x + 9.792
Now, to find the value of x, we can use numerical methods such as the Newton-Raphson method or graphical methods such as plotting the equation and finding its roots.
Once you have the value of x, you can use it to calculate the moles of each substance at equilibrium.
For example, the moles of N2 at equilibrium would be 1 - x, the moles of H2 would be 4 - 3x, and the moles of NH3 would be 2x.