Question

For the reaction \( \text{N}_2(g) + 3\text{H_2(g) \rightleftharpoons 2\text{NH}_3(g) \), the equilibrium constant \( K_c \) at a certain temperature is 1.5. If the concentration of \( \text{N}_2 \) is 0.5 M, \( \text{H}_2 \) is 1.0 M, and \( \text{NH}_3 \) is 0.2 M, what is the reaction quotient \( Q_c \)?}

• A. \( 1.0 \)
• B. \( 1.5 \)
• C. \( 0.5 \)
• D. \( 2.0 \)

Hint:

Remember that \( Q_c \) can be calculated in the same way as the equilibrium constant, using the concentrations of the reactants and products at any point during the reaction.

Answer 1

The Correct Option is C

Step 1: Understand the reaction quotient. The reaction quotient \( Q_c \) for a chemical reaction is calculated using the concentrations of the products and reactants, and is given by the following expression for the reaction \( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \): \[ Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \] Step 2: Substitute the given values. We are given: - \( [\text{N}_2] = 0.5 \, \text{M} \), - \( [\text{H}_2] = 1.0 \, \text{M} \), - \( [\text{NH}_3] = 0.2 \, \text{M} \). Substitute these values into the equation for \( Q_c \): \[ Q_c = \frac{(0.2)^2}{(0.5)(1.0)^3} \] \[ Q_c = \frac{0.04}{0.5} = 0.08 \] Answer: Therefore, the value of \( Q_c \) is \( 0.08 \), which is not one of the given options, so there may have been a typo in the options provided. However, based on the calculation, this is the correct result.