4AB › Chemistry PYQ › CSIR-NET › June 2016 › Q56Q56 · CSIR-NET Chemistry, June 2016 Paper: CSIR-NET June 2016 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Rate Laws Order · Marks: 2 · Difficulty: Hard
If the rates of a reaction are $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ at concentrations $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ of a reactant respectively, the order of reaction, 'n' (assuming that the concentrations of all other reactants and T remain constant) with respect to that reactant is given by
(a) $\mathrm{n}=\frac{\log \mathrm{R}_{1}-\log \mathrm{R}_{2}}{\log \mathrm{C}_{1}-\log \mathrm{C}_{2}}$
(b) $\mathrm{n}=\frac{\log \mathrm{C}_{1}-\log \mathrm{C}_{2}}{\log \mathrm{R}_{1}-\log \mathrm{R}_{2}}$
(c) $\mathrm{n}=\frac{\log \mathrm{C}_{1}-\log \mathrm{R}_{1}}{\log \mathrm{C}_{2}-\log \mathrm{R}_{2}}$
(d) $\mathrm{n}=\frac{\log \mathrm{C}_{2}-\log \mathrm{R}_{2}}{\log \mathrm{C}_{1}-\log \mathrm{R}_{1}}$
Answer Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation R = kCⁿ ⇒ log R₁ − log R₂ = n(log C₁ − log C₂), so n = (log R₁ − log R₂)/(log C₁ − log C₂).
← Q55 · All questions in this paper · Chemical Kinetics chapter list · Q57 → Study loop for Chemical Kinetics
Open in whiteboard · Browse this chapter in the app