Question

For the reaction \(2\text{NOBr}_{(g)}\rightarrow 2\text{NO}_{(g)}+\text{Br}_{2(g)}\) rate law is \(r = k[\text{NOBr}]^2\), if rate constant is \(1\cdot 62\text{ mol dm}^{-3}\text{ s}^{-1}\) and concentration of NOBr is \(2\times 10^{-3}\text{ mol L}^{-1}\). What is the rate of reaction ?

• A. \(4\cdot 05\times 10^{-5}\text{ mol dm}^{-3}\text{s}^{-1}\)
• B. \(6\cdot 48\times 10^{-6}\text{ mol dm}^{-3}\text{s}^{-1}\)
• C. \(2\cdot 12\times 10^{-5}\text{ mol dm}^{-3}\text{s}^{-1}\)
• D. \(8\cdot 63\times 10^{-6}\text{ mol dm}^{-3}\text{s}^{-1}\)

Hint:

Convert the concentration to the same units as k and substitute into the rate law.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The rate law is \(r = k[\text{NOBr}]^2\). The unit of k is mol dm\(^{-3}\) s\(^{-1}\) as given, and \(1\text{ L} = 1\text{ dm}^3\), so the concentration \(2\times10^{-3}\) mol L\(^{-1}\) is \(2\times10^{-3}\) mol dm\(^{-3}\).

Step 2: Calculation:
\[ r = 1.62\times(2\times 10^{-3})^2 = 1.62\times 4\times 10^{-6} = 6.48\times 10^{-6}\text{ mol dm}^{-3}\text{ s}^{-1} \]
The other options come from errors in squaring (for example, not squaring the concentration gives \(3.24\times10^{-3}\), which is not listed).

Final Answer:
The rate of the reaction is \(6.48\times 10^{-6}\) mol dm\(^{-3}\) s\(^{-1}\), option (B). \[ \boxed{6.48\times 10^{-6}\text{ mol dm}^{-3}\text{s}^{-1}} \]