Answered

Consider the reaction:

[tex]\[
N_2(g) + 3 H_2(g) \longleftrightarrow 2 NH_3(g)
\][/tex]

At equilibrium, the concentrations of the different species are as follows:

[tex]\[
\begin{array}{l}
[NH_3] = 0.105 \, M \\
[N_2] = 1.1 \, M \\
[H_2] = 1.50 \, M
\end{array}
\][/tex]

What is the equilibrium constant for the reaction at this temperature?

A. 0.0030
B. 0.030
C. 34
D. 340



Answer :

To find the equilibrium constant \( K_c \) for the given reaction at equilibrium, we can use the equilibrium concentrations given and the formula for the equilibrium constant for the reaction:

[tex]\[ \text{N}_2(g) + 3 \text{H}_2(g) \longleftrightarrow 2 \text{NH}_3(g) \][/tex]

The equilibrium constant expression for this reaction is:

[tex]\[ K_c = \frac{{[\text{NH}_3]^2}}{{[\text{N}_2] \cdot [\text{H}_2]^3}} \][/tex]

Given:
[tex]\[ [\text{NH}_3] = 0.105 \, M \][/tex]
[tex]\[ [\text{N}_2] = 1.1 \, M \][/tex]
[tex]\[ [\text{H}_2] = 1.50 \, M \][/tex]

We substitute these values into the equilibrium constant expression:

[tex]\[ K_c = \frac{{(0.105)^2}}{{(1.1) \cdot (1.50)^3}} \][/tex]

Evaluating the numerator and the denominator:

[tex]\[ \text{Numerator} = (0.105)^2 = 0.011025 \][/tex]

[tex]\[ \text{Denominator} = 1.1 \cdot (1.50)^3 = 1.1 \cdot 3.375 = 3.7125 \][/tex]

Now, divide the numerator by the denominator:

[tex]\[ K_c = \frac{0.011025}{3.7125} \approx 0.0029697 \][/tex]

Rounding this to three significant figures, we get:

[tex]\[ K_c \approx 0.0030 \][/tex]

Therefore, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction at this temperature is 0.0030.