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For the following reaction $A (g) + B (g) ⟶ 2 C (g) + D (g)$ , the rate law is given as $R = K [A]^{1 / 3} \cdot [B]^{2 / 3}$ If initially concentration of A and B are unity then total molar concentration of a11 gases at t c 600 sec. will be: [Given: $K = 2.31 \times 10^{- 3} \sec^{- 1}$ ]

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1 mole/litre

l .25 mole/Iitre

3.5 mole/litre

2. 75 mole/Iitre

answer is D.

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Detailed Solution

$A + B ⟶ 2 C + D$

Concentration for A and B is 1 for each at t=0.

Concentration for A and B is 1-x for each, for C it is 2x and for D it is x at t=600.

So,

$\begin{array}{l} \frac{d x}{d t} = k (1 - x) \\ k \times t = \ln [\frac{1}{1 - x}] \\ 2.31 \times 10^{- 3} \times 600 = \ln [\frac{1}{1 - x}] \\ \Rightarrow = \frac{3}{4} \end{array}$

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