Exams › JEE Main › Chemistry
For the reaction between nitric oxide and bromine forming nitrosyl bromide, the following sequence is suggested: NO(g) + Br2(g) → NOBr2(g) NOBr2(g) + NO(g) → 2NOBr(g) If the second step controls the rate, what is the reaction order with respect to NO(g)?
- 3
- 2
- 1
- 0
Correct answer: 2
Solution
The reaction order with respect to NO(g) is 2 because the rate-determining step involves two molecules of NO(g) reacting with NOBr2(g), indicating that the concentration of NO(g) is squared in the rate law.
Related JEE Main Chemistry questions
- If a biochemical reaction is performed in the laboratory without an enzyme and its rate is found to be 10^{-6} times, then the activation energy for the reaction in the presence of the enzyme is:
- For the reaction A \rightarrow products, the half-life is 120 min when the initial concentration of A is 8.0 \times 10^{-2} M. If the starting concentration is changed to 4.0 \times 10^{-2} M, the half-life increases to 240 min. What is the reaction order?
- For a first-order process A \rightarrow B, if the rate constant is k and the initial concentration of A is 0.5 M, what is the half-life?
- For the gas-phase reaction A(g) + B(g) \rightarrow C(g), the rate law is rate = k C_A^2 C_B^4. Which change in the initial concentrations of A and B will make the reaction rate eight times larger?
- For the elementary gas-phase reaction A(g) + 2B(g) \rightarrow C(g) + D(g), an experiment starts with initial partial pressures p_A = 0.60 atm and p_B = 0.80 atm. If the partial pressure of C becomes 0.20 atm, then the reaction rate compared with the initial rate is
- From the experimental data below for the reaction A + B \rightarrow products, identify the rate law that fits the observations. Exp. 1: [A] = 0.012, [B] = 0.035, initial rate = 0.1 Exp. 2: [A] = 0.024, [B] = 0.070, initial rate = 0.8 Exp. 3: [A] = 0.024, [B] = 0.035, initial rate = 0.1 Exp. 4: [A] = 0.012, [B] = 0.070, initial rate = 0.8
⚔️ Practice JEE Main Chemistry free + battle 1v1 →