Given the rate law for a reaction is rate = k[NO][O_2]^2 and the initial rate is 0.020 mol L^{-1} s^{-1} when [NO] = 0.10 mol L^{-1} and [O_2] = 0.20 mol L^{-1}, what is the value of the rate constant k at this temperature?

• A. 5.0 mol^{-2} L^{2} s^{-1}
• B. 0.10 mol^{-2} L^{2} s^{-1}
• C. 2.0 mol^{-2} L^{2} s^{-1}
• D. 0.50 mol^{-2} L^{2} s^{-1}

Answer: (A)

Verified step by step guidance

1. Write down the given rate law: \(\text{rate} = k[\text{NO}][\text{O}_2]^2\).
2. Identify the known values from the problem: rate = 0.020 mol L\(^{-1}\) s\(^{-1}\), \([\text{NO}] = 0.10\) mol L\(^{-1}\), and \([\text{O}_2] = 0.20\) mol L\(^{-1}\).
3. Substitute the known values into the rate law equation: \$0.020 = k \times 0.10 \times (0.20)^2$.
4. Simplify the expression on the right side by calculating the concentration terms: \$0.10 \times (0.20)^2 = 0.10 \times 0.04$.
5. Solve for the rate constant \(k\) by dividing both sides of the equation by the product of the concentrations: \(k = \frac{0.020}{0.10 \times 0.04}\).