Question

The following concentrations were obtained in the formation of $NH_{3(g)}$ from $N_{2(g)}$ and $H_{2(g)}$ at equilibrium at $500\ \text{K}$: $[NH_{3}]=1.5\times10^{-2}\ \text{M}$, $[N_{2}]=5\times10^{-3}\ \text{M}$ and $[H_{2}]=0.10\ \text{M}$. Calculate the equilibrium constant $K_{c}$ for the reaction $N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.

• A. 0.45
• B. 4.5
• C. 45.0
• D. $4.5\times10^{-2}$
• E. $4.5\times10^{-3}$

Hint:

Square the product concentration and cube the $H_{2}$ concentration per the coefficients.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem requires the calculation of the equilibrium constant, \( K_c \), for a chemical reaction given the equilibrium concentrations of the reactants and products.
Step 2: Key Formula or Approach:
For a general reversible reaction: \[ aA + bB \rightleftharpoons cC + dD \] The equilibrium constant in terms of concentration (\( K_c \)) is given by the expression: \[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \] where [X] represents the molar concentration of species X at equilibrium. The unit conversion note "dm\(^6\) mol\(^{-2}\)" is consistent with the units of \(K_c\) for this reaction, as \( 1 \text{ dm}^3 = 1 \text{ L} \), and molarity (M) is in mol/L. Step 3: Detailed Explanation:
The given balanced reaction is: \[ \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \] The expression for the equilibrium constant \( K_c \) is: \[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \] We are given the equilibrium concentrations: - [NH\(_3\)] = \( 1.5 \times 10^{-2} \) M - [N\(_2\)] = \( 5 \times 10^{-3} \) M - [H\(_2\)] = 0.10 M = \( 1 \times 10^{-1} \) M Substitute these values into the \( K_c \) expression: \[ K_c = \frac{(1.5 \times 10^{-2})^2}{(5 \times 10^{-3}) (0.10)^3} \] Calculate the numerator: \[ (1.5 \times 10^{-2})^2 = 1.5^2 \times (10^{-2})^2 = 2.25 \times 10^{-4} \] Calculate the denominator: \[ (5 \times 10^{-3}) \times (1 \times 10^{-1})^3 = (5 \times 10^{-3}) \times (10^{-3}) = 5 \times 10^{-6} \] Now, divide the numerator by the denominator: \[ K_c = \frac{2.25 \times 10^{-4}}{5 \times 10^{-6}} \] \[ K_c = \frac{2.25}{5} \times 10^{-4 - (-6)} = 0.45 \times 10^2 = 45.0 \] The units are \( \frac{M^2}{M \cdot M^3} = M^{-2} = (\text{mol/L})^{-2} = \text{L}^2\text{mol}^{-2} = \text{dm}^6\text{mol}^{-2} \). Step 4: Final Answer:
The equilibrium constant \( K_c \) is 45.0.