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CBSE Class 12 Chemistry 2023 All Sets Solved Paper

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Question : 13 of 18
Marks: +1, -0
The following experimental rate data were obtained for a reaction carried out at 25∘C:25^{\circ} \mathrm{C} :
A(g)+B(g)→C(g)+D(g)A_{(g)}+B_{(g)}\rightarrow C_{(g)}+D_{(g)}
Initial [A(g)]mol dm−3\frac{[A_{(g)}]}{\mathrm{mol}\,\mathrm{dm}^{-3}}Initial [B(g)]mol dm−3\frac{[B_{(g)}]}{\mathrm{mol}\,\mathrm{dm}^{-3}}Initial rate/mol dm−3 s−1\mathrm{dm}^{-3}\,\mathrm{s}^{-1}
3.0×10−23.0 \times 10^{-2}2.0×10−22.0 \times 10^{-2}1.89×10−41.89 \times 10^{-4}
3.0×10−23.0 \times 10^{-2}4.0×10−24.0 \times 10^{-2}1.89×10−41.89 \times 10^{-4}
6.0×10−26.0 \times 10^{-2}4.0×10−24.0 \times 10^{-2}7.56×10−47.56 \times 10^{-4}
What are the orders with respect to A(g)A_{(g)} and B(g)B_{(g)} ?
Solution:  
Let say, rate of reaction = K[A]x[B]yK[A]^x[B]^y.
then,
1.89×10−41.89 \times 10^{-4} = K[3.0×10−2]x[2×10−2]yK[3.0 \times 10^{-2}]^x[2 \times 10^{-2}]^y _____(i)
1.89×10−4=K[3×10−2]x[4×10−2]y1.89 \times 10^{-4} = K[3 \times 10^{-2}]^x[4 \times 10^{-2}]^y _____(ii)
and 7.56×10−4=K[6×10−2]x[4×10−2]y7.56 \times 10^{-4} = K[6 \times 10^{-2}]^x[4 \times 10^{-2}]^y _____(iii)
from equation (i) and (ii) , [ii÷i][\text{ii} \div \text{i}] we got
1=2y1 = 2^y
⇒2∘=2y\Rightarrow 2^{\circ} = 2^y
⇒y=0\Rightarrow y = 0
from equation [iii÷ii][\text{iii} \div \text{ii}], we got
4=2x4 = 2^x
22=2x2^2 = 2^x
x=2x = 2
therefore, the order of reaction with respect to A = 2 and w.r.to B = O
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