Question

For the reaction \(2\text{NOBr}_{(g)}⟶2\text{NO}_{(g)}+\text{Br}_{2(g)}\), rate law is \(r = k[\text{NOBr}]^2\). Rate constant is \(1.62 \text{M s}^{-1}\) and concentration of NOBr is \(5\times 10^{-3} \text{M}\),
What is rate of reaction?

• A. \(4.05\times 10^{-5} \text{M s}^{-1}\)
• B. \(6.48\times 10^{-6} \text{M s}^{-1}\)
• C. \(2.12\times 10^{-5} \text{M s}^{-1}\)
• D. \(8.63\times 10^{-6} \text{M s}^{-1}\)

Hint:

Substitute concentration into rate = k[NOBr]^2.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The rate law \(r = k[\text{NOBr}]^2\) says the rate depends on the square of the concentration. We just substitute the numbers.

Step 2: Key Formula or Approach:
\(r = k[\text{NOBr}]^2\) with \(k = 1.62\) and \([\text{NOBr}] = 5 \times 10^{-3}\) M.

Step 3: Detailed Explanation:
\([\text{NOBr}]^2 = (5 \times 10^{-3})^2 = 25 \times 10^{-6} = 2.5 \times 10^{-5}\).
\[ r = 1.62 \times 2.5 \times 10^{-5} = 4.05 \times 10^{-5}\ \text{M s}^{-1} \]
If the concentration was not squared we would get \(8.1 \times 10^{-3}\), which is not among the options. The other values do not follow from the given numbers.

Final Answer:
The rate is \(4.05 \times 10^{-5}\) M/s, option (A). \[ \boxed{4.05 \times 10^{-5}\ \text{M s}^{-1}} \]