Skip to content
dronacharya.ai
← JEE Advanced 2025 · Paper 2
ChemistryNumerical+4Question 44 of 48

JEE Advanced 2025 · Paper 2, Chemistry question 12

Consider a reaction A+R⟶ProductA + R \longrightarrow \textit{Product}. The rate of this reaction is measured to be k[A][R]k[A][R]. At the start of the reaction, the concentration of RR, [R]0[R]_0, is 10-times the concentration of AA, [A]0[A]_0. The reaction can be considered to be a pseudo first order reaction with assumption that k[R]=k′k[R] = k' is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40 % complete, is _____.

[kk and k′k' represent corresponding rate constants]

Answer and worked solution

Official answer: 4

At 40 % completion:

  • [A]=0.6[A]0[A]=0.6[A]_0;
  • [R]=[R]0−0.4[A]0=9.6[A]0[R]=[R]_0-0.4[A]_0=9.6[A]_0.

True rate. k(0.6[A]0)(9.6[A]0)k(0.6[A]_0)(9.6[A]_0).

Pseudo-first-order rate. k(10[A]0)(0.6[A]0)k(10[A]_0)(0.6[A]_0), since it keeps [R]=[R]0[R]=[R]_0.

Relative error.
10−9.69.6×100≈4.17 %.\frac{10-9.6}{9.6}\times100\approx4.17\ \%.