Test Index

NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

© examsnet.com
Question : 12 of 39
Marks: +1, -0
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table :
 Experiment  [A]mol L−1\frac{[A]}{\text{mol}\,\text{L}^{-1}}  [B]mol L−1\frac{[B]}{\text{mol}\,\text{L}^{-1}}  Initial rate/mol L−1 min−1\text{L}^{-1}\,\text{min}^{-1}
 I  0.1  0.1  2.0×10−22.0 \times 10^{-2}
 II  â€“  0.2  4.0×10−24.0 \times 10^{-2}
 III  0.4  0.4  â€“
 IV   –  0.2  2.0×10−22.0 \times 10^{-2}
Solution:  
The reaction is first order with respect to A and zero order withrespect to B, hence the rate expression will be =k[A]1[B]0=k[A]=k[A]^{1}[B]^{0}=k[A]
From expt. I: 2.0×10−2=k(0.1)2.0 \times 10^{-2}=k(0.1)
or, k=0.2 min−1k=0.2 \text{ min}^{-1}
From expt. II: 4.0×10−2=(0.2)×[A]4.0 \times 10^{-2}=(0.2) \times [A]
⇒[A]=0.2 mol L−1\Rightarrow [A]=0.2 \text{ mol L}^{-1}
From expt. III: Rate =(0.2)(0.4)=0.08 mol L−1 min−1=(0.2)(0.4)=0.08 \text{ mol L}^{-1} \text{ min}^{-1}
=8×10−2 mol L−1 min−1=8 \times 10^{-2} \text{ mol L}^{-1} \text{ min}^{-1}
From expt. IV: 2.0×10−2=(0.2)×[A]2.0 \times 10^{-2}=(0.2) \times [A]
⇒[A]=0.1 mol L−1\Rightarrow [A]=0.1 \text{ mol L}^{-1}
© examsnet.com
Go to Question:
Prev Question
Next Question