Hi this question is on my study guide and I want to make sure I get to the right answer and know why.
Calculate K for the synthesis of ammonia at 400°C
N2(g) + 3 H2(g) —> 2 NH3(g)
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, if the following concentrations are present at equilibrium: N2(g)= 1.2 mol/L, H2(g) = 0.80 mol/L, and NH3(g) = 0.28 mol/L.
(Subscript 2 in N2, 3 H2, and subscript 3 in 2 NH3, double arrows in reaction)
Write the equilibrium K expression.
Substitute the numbers for the concentrations shown at equilibrium.
Calculate K.
I'll be glad to check your work if you post it.
K = (0.28)^2 / 1.2 ( 0.80)^3 =1.28 x 10^-1 ?
I agree. Good work. If your prof is picky about significant figures, s/he will object to using 3 s.f. in the answer when you have only 2 in the numbers; therefore, you may want to round that final answer to 0.13
Thanks
To calculate the equilibrium constant (K) for the synthesis of ammonia, you need to use the formula:
K = [NH3]^2 / ([N2] * [H2]^3)
In the given reaction, you are given the concentrations at equilibrium:
[N2(g)] = 1.2 mol/L
[H2(g)] = 0.80 mol/L
[NH3(g)] = 0.28 mol/L
Substituting the given values into the equilibrium constant expression, you get:
K = (0.28 mol/L)^2 / ((1.2 mol/L) * (0.80 mol/L)^3)
Simplifying:
K = 0.784 / (0.96 * 0.512)
K = 0.784 / 0.49152
K ≈ 1.595
So, the equilibrium constant (K) for the synthesis of ammonia at 400°C is approximately 1.595.