JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The results given in the below table were obtained during kinetic studies of the following reaction: \(2 A + B \longrightarrow C + D\)
| Experiment | \([ A ] / molL ^{-1}\) | \([ B ] / molL ^{-1}\) | Initial \(rate/molL\) \(^{-1}\) \(\min ^{-1}\) |
| \(I\) | \(0.1\) | \(0.1\) | \(6.00 \times 10^{-3}\) |
| \(II\) | \(0.1\) | \(0.2\) | \(2.40 \times 10^{-2}\) |
| \(III\) | \(0.2\) | \(0.1\) | \(1.20 \times 10^{-2}\) |
| \(IV\) | \(X\) | \(0.2\) | \(7.20 \times 10^{-2}\) |
| \(V\) | \(0.3\) | \(Y\) | \(2.88 \times 10^{-1}\) |
- A \(0.3,0.4\)
- B \(0.4,0.3\)
- C \(0.4,0.4\)
- D \(0.3,0.3\)
Answer & Solution
Correct Answer
(A) \(0.3,0.4\)
Step-by-step Solution
Detailed explanation
From rate law \(= K [ A ]^{ x }[ B ]^{ y }\) \(6 \times 10^{-3}= K (0.1)^{ x }(0.1)^{ y }\)\(.....(1)\) \(2.4 \times 10^{-2}= K (0.1)^{ x }(0.2)^{ y } \ldots \ldots(2)\) \(1.2 \times 10^{-2}= K (0.2)^{ x }(0.1)^{ y }\)\(......(3)\) \((3) \div(1) \Rightarrow x=1\)…
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