Given the equilibrium concentrations in the table, what is the equilibrium constant for the synthesis of ammonia at this temperature?
3H2(g) + N2 <---> 2NH3(g)
[H2] - 0.5
[N2] - 0.23
[NH3] - 0.020
A. 0.0035
B. 0.014
C. 0.066
D. 0.017
Keq=[NH3]^2/[H2]^2[N2]. Plug in the concentrations for each and you should get 0.014. So the answer is B.
To calculate the equilibrium constant (K) for the synthesis of ammonia, we need to use the formula:
K = ([NH3]^2) / ([H2]^3 * [N2])
According to the given equilibrium concentrations:
[H2] = 0.5
[N2] = 0.23
[NH3] = 0.020
Now we can substitute these values into the formula:
K = (0.020^2) / (0.5^3 * 0.23)
K = 0.0004 / (0.125 * 0.23)
K = 0.0004 / 0.02875
K ≈ 0.0139
The closest answer choice to 0.0139 is option B. Therefore, the equilibrium constant for the synthesis of ammonia at this temperature is approximately 0.014.
To calculate the equilibrium constant (K) for the synthesis of ammonia at a given temperature, you need to use the equilibrium concentrations of the reactants and products. The equilibrium constant expression for the reaction is:
K = ([NH3]^2) / ([H2]^3 * [N2])
In this case, the equilibrium concentrations given are [H2] = 0.5, [N2] = 0.23, and [NH3] = 0.020.
Plugging these values into the equation:
K = (0.020^2) / (0.5^3 * 0.23)
Now, to obtain the value of K, perform the calculations:
K = 0.0004 / 0.0575
K ≈ 0.0069565
Now, compare this value to the options provided:
A. 0.0035
B. 0.014
C. 0.066
D. 0.017
The closest option to the calculated value is option B, 0.014. Therefore, the equilibrium constant for the synthesis of ammonia at this temperature is approximately 0.014.