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JEE Advanced · Chemistry · 18. Chemical Kinetics

Consider the kinetic data given in the following table for the reaction A+B+C Product.
Ex
peri
ment No.
\([\text A]\) \((\text{mol}\) \(\text{dm} ^{-3})\)\([\text B]\) \((\text{mol}\) \(\text{dm} ^{-3})\)\([\text C]\) \((\text{mol}\) \(\text{dm} ^{-3})\)Rate
of
reaction \((\text{mol}\) \(\text{dm} ^{-3}\) \(\text s^{-1})\)
10.20.10.1\(6.0 \times\) \(10^{-5}\)
20.20.20.1\(6.0 \times \) \(10^{-5}\)
30.20.10.2\(1.2 \times \) \(10^{-4}\)
40.30.10.1\(9.0 \times \) \(10^{-5}\)
The rate of the reaction for A=0.15 mol dm-3, B=0.25 mol dm-3 and C=0.15 mol dm-3 is found to be Y×10-5 mol dm-3s-1. The value of Y is __________

  1. A 6.75
  2. B 7.56
  3. C 10.5
  4. D 11.89
Verified Solution

Answer & Solution

Correct Answer

(A) 6.75

Step-by-step Solution

Detailed explanation

r=k[A] a [B]b [C]c
6×10-5=k(0.2)a (0.1)b (0.1)c ..... (1)
6×10-5=k(0.2)a(0.2)b(0.1)c ...... (2)
1.2×10-4=k(0.2)a(0.1)b (0.2)c ...... (3)
9×10-5=k(0.3)a(0.1)b (0.1)c ...... (4)
From equation (1) and (ii) : b=0
From equation (1) and (iii) : c=1
From equation (1) and (iv) : a=1
Put x,y and z in equation (i) we get;
6×10-5=k(0.2)1 (0.1)0(0.1)1
r=3×10-3[A][C]
For [A]=0.15 mol dm-3
[B]=0.25 mol dm-3
[C]=0.15 mol dm-3
r=3×10-3×0.15×0.15
r=6.75×10-5
r=6.75
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