Chemical Kinetics and Nuclear ChemistryHard

Question

For a reaction 2A + B + 3C → D + 3E, the following data is obtained.

Exp. No. Concentration in mole per litre Initial rate of formation of D (torr s−1)

A B C

1 0.01 0.01 0.01 2.5 × 10−4

2 0.02 0.01 0.01 1.0 × 10−3

3 0.01 0.02 0.01 2.5 × 10−4

4 0.01 0.02 0.02 5.0 × 10−4

The order with respect to A, B and C are, respectively,

Options

A.0, 1, 2
B.2, 0, 1
C.1, 0, 2
D.2, 1, 1

Solution

r = K[A]x [B]y [C]z

Exp (1) ÷ Exp (2): $\frac{2.5 \times 10^{- 4}}{1.0 \times 10^{- 3}} = \frac{K(0.01)^{x}(0.01)^{y}(0.01)^{z}}{K(0.02)^{x}(0.01)^{y}(0.01)^{z}} \Rightarrow x = 2$

Exp (1) ÷ Exp (3): $\frac{2.5 \times 10^{- 4}}{2.5 \times 10^{- 4}} = \frac{K(0.01)^{x}(0.01)^{y}(0.01)^{z}}{K(0.01)^{x}(0.02)^{y}(0.01)^{z}} \Rightarrow y = 0$

Exp (3) ÷ Exp (4): $\frac{2.5 \times 10^{- 4}}{5.0 \times 10^{- 4}} = \frac{K(0.01)^{x}(0.02)^{y}(0.01)^{z}}{K(0.01)^{x}(0.02)^{y}(0.02)^{z}} \Rightarrow z = 1$

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