AP EAMCET · Chemistry · Chemical Kinetics
The rate equation for the reaction \(2 A+B \longrightarrow\) products is rate \(=k[A][B]^2\).
If \(k\) at \(T(\mathrm{~K})\) is \(5.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1}\), the initial rate of the reaction, when \([A]=0.05 \mathrm{~mol} \mathrm{~L}^{-1}\) and \([B]=0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) is
- A \(1.25 \times 10^{-9} \mathrm{~L} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
- B \(1.25 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
- C \(2.50 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
- D \(2.50 \times 10^{-9} \mathrm{~L} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(C) \(2.50 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
For the reaction, \(2 A+B \longrightarrow\) Products \(\text {Rate }=k[A][B]^2\) Given, Rate constant, \(k\) of reaction \(=5.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1}\)…
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