Ncert Exercises & Solutions

4.2 For the reaction:

2A + B → A2B

the rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1.

NEETprep Answer:

The initial rate of the reaction is

Rate o=K $[A] {[B]}^{2}$
$= (2.0 \times 10^{- 6} m o l^{- 2} L^{2} s^{- 1}) (0.1 m o l L^{- 1}) (0.2 m o l L$ ^-1)2
^{=8.0 $\times 10^{- 9} m o l L^{2} s^{- 1}$}

When [A] is reduced from 0.1 mol L $^{- 1}$ to 0.06 mol $^{- 1}$ ,the concentration of A reacted=(0.1-0.6) mol $L - 1 = 0.04 m o l L^{- 1}$

therefore, concentration of B reacted= $\frac{1}{2} \times 0.04 m o l L^{- 1} = 0.02 m o l L^{- 1}$

Then, concentration of B available, ,[B] =(0.2-0.2) mol $L^{1}$

=0.18 mol $L^{1}$

After [A] is reduced to 0.06 mol L, the rate of the reaction is given by,

Rate= K $[A] {[B]}^{2}$

=(2.0 $\times$ 10 $^{- 6} m o l^{- 2} L^{2} s^{- 1}) (0.06 m o l L^{- 1}) (0.18 m o l L^{- 1})$
$= 3.89 m o l L^{- 1} s^{- 1}$

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