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Calculate Relative Error in Rate for Pseudo First Order Reaction

Consider a reaction A+RProductA + R \rightarrow 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 1010-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]=kk[R] = k' is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40%40\% complete, is ______.

[kk and kk' represent corresponding rate constants]

Official Numerical Answer4 to 4.25

Step-by-Step Solution

To find the relative error in the rate of reaction, we evaluate both the actual rate and the assumed rate under pseudo-first-order conditions when the reaction is 40%40\% complete.

Step 1: Initial Conditions and Concentrations at 40%40\% Completion

The given elementary reaction is: A+RProductA + R \rightarrow \text{Product}

The actual rate law is given by: ractual=k[A][R]r_{\text{actual}} = k[A][R]

At the start of the reaction (t=0t = 0):

  • [A]=[A]0[A] = [A]_0
  • [R]0=10[A]0[R]_0 = 10[A]_0

When the reaction is 40%40\% complete, 40%40\% of reactant AA has reacted:

  • Remaining concentration of AA: [A]=[A]00.40[A]0=0.60[A]0[A] = [A]_0 - 0.40[A]_0 = 0.60[A]_0

Since 1 mole of AA reacts with 1 mole of RR, the amount of RR consumed is also 0.40[A]00.40[A]_0:

  • Remaining concentration of RR: [R]=[R]00.40[A]0=10[A]00.40[A]0=9.60[A]0[R] = [R]_0 - 0.40[A]_0 = 10[A]_0 - 0.40[A]_0 = 9.60[A]_0

Step 2: Calculation of Actual and Assumed Rates

  1. Actual Rate (ractualr_{\text{actual}}): ractual=k[A][R]=k(0.60[A]0)(9.60[A]0)=5.76k[A]02r_{\text{actual}} = k[A][R] = k(0.60[A]_0)(9.60[A]_0) = 5.76 k[A]_0^2

  2. Assumed Pseudo-First-Order Rate (rassumedr_{\text{assumed}}): Under the assumption that k[R]=k=k[R]0k[R] = k' = k[R]_0 is constant throughout the reaction: k=k[R]0=10k[A]0k' = k[R]_0 = 10 k[A]_0

    Thus, the assumed rate is: rassumed=k[A]=(10k[A]0)(0.60[A]0)=6.00k[A]02r_{\text{assumed}} = k'[A] = (10 k[A]_0)(0.60[A]_0) = 6.00 k[A]_0^2


Step 3: Calculation of Relative Error

The difference in the rate due to the assumption is: Δr=rassumedractual=6.00k[A]025.76k[A]02=0.24k[A]02\Delta r = r_{\text{assumed}} - r_{\text{actual}} = 6.00 k[A]_0^2 - 5.76 k[A]_0^2 = 0.24 k[A]_0^2

  1. Relative Error with respect to Actual Rate: Relative Error (%)=rassumedractualractual×100%=0.24k[A]025.76k[A]02×100%=124×100%4.17%\text{Relative Error (\%)} = \frac{r_{\text{assumed}} - r_{\text{actual}}}{r_{\text{actual}}} \times 100\% = \frac{0.24 k[A]_0^2}{5.76 k[A]_0^2} \times 100\% = \frac{1}{24} \times 100\% \approx 4.17\%

  2. Relative Error with respect to Assumed Rate: Relative Error (%)=rassumedractualrassumed×100%=0.24k[A]026.00k[A]02×100%=4.00%\text{Relative Error (\%)} = \frac{r_{\text{assumed}} - r_{\text{actual}}}{r_{\text{assumed}}} \times 100\% = \frac{0.24 k[A]_0^2}{6.00 k[A]_0^2} \times 100\% = 4.00\%

Both 4.17%4.17\% and 4.00%4.00\% fall within the standard acceptable range [4.0,4.25][4.0, 4.25].

Final Answer: The relative error (in %) in the rate is 4.17 (or 4).

Calculate Relative Error in Rate for Pseudo First Order Reaction | Chemistry PYQ Solution - JEE Challenger