At 850 K, the value of the equilibrium constant Kp for the ammonia synthesis reaction N2(g) + H2(g) <--> N2H2(g) is 0.1690. If a vessel contains an initial reaction mixture in which [N2]=0.0150 M, [H2]=0.0200 M, and [N2H2]=0.000250 M, what will the [N2H2] be when equilibrium is reached?
You didn't balance th equation and ammonia is NH3. N2H4 is hydrazine.
Did you intend to write concns (and not partial pressures) for the reactants and products.
I think so. That is what the question said.
What about the product? Is that N2H4 or NH3.
the equation was given to me as is, along with the initial concentrations
To find the equilibrium concentration of N2H2, we will need to use the equilibrium expression and make use of the given equilibrium constant. Here's how we can calculate it:
Step 1: Write the balanced equation for the reaction:
N2(g) + H2(g) <--> N2H2(g)
Step 2: Write the equilibrium expression:
Kp = [N2H2] / ([N2]*[H2])
Step 3: Plug in the given values:
Kp = 0.1690
[N2] = 0.0150 M
[H2] = 0.0200 M
Step 4: Solve for [N2H2]:
[N2H2] = Kp * ([N2] * [H2])
[N2H2] = 0.1690 * (0.0150 M * 0.0200 M)
[N2H2] = 0.00005085 M
So, the concentration of N2H2 when equilibrium is reached is 0.00005085 M.