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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
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Question : 2 of 39
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For the reaction : 2A + B → A2_{2}B the rate=k[A][B]2= k[A][B]^{2} with k =2.0×10−6 mol−2 L2 s−1.= 2.0 \times 10^{-6}\ \mathrm{mol}^{-2}\ \mathrm{L}^{2}\ \mathrm{s}^{-1}. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1^{-1}, [B] = 0.2 mol L−1^{-1}. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1^{-1}.
Solution:  
Initial rate =k[A][B]2=k[A][B]^{2} =(2.0×10−6)×(0.1)×(0.2)2= (2.0 \times 10^{-6}) \times (0.1) \times (0.2)^{2} =8×10−9 mol L−1 s−1=8 \times 10^{-9}\ \mathrm{mol}\ \mathrm{L}^{-1}\ \mathrm{s}^{-1}
When [A] is reduced from 0.10 mol L−10.10\ \mathrm{mol}\ \mathrm{L}^{-1} to 0.06 mol L−10.06\ \mathrm{mol}\ \mathrm{L}^{-1} , i.e. 0.04 mol L−1\text{i.e.}\ 0.04\ \mathrm{mol}\ \mathrm{L}^{-1} of A has reacted. Then according to equation, amount of B reacted is half of A.
[B]=12×0.04[B]=\frac{1}{2} \times 0.04 mol L−1\mathrm{L}^{-1} =0.02=0.02mol L−1\mathrm{L}^{-1}
Hence, [B] =0.2−0.02=0.18 mol L−1=0.2-0.02=0.18\ \mathrm{mol}\ \mathrm{L}^{-1}
Now, again rate =k[A][B]2=k[A][B]^{2}
∴  \therefore\; Rate =(2.0×10−6)×(0.06)×(0.18)2= (2.0 \times 10^{-6}) \times (0.06) \times (0.18)^{2} =3.89×10−9 mol L−1 s−1=3.89 \times 10^{-9}\ \mathrm{mol}\ \mathrm{L}^{-1}\ \mathrm{s}^{-1}
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