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Question: For reaction A + 2B \(\longrightarrow\)C + D rate law R = k[A]<sup>1</sup>[B]<sup>2</sup>. By what ...

For reaction A + 2B ⟶\longrightarrowC + D rate law

R = k[A]1[B]2. By what factor would the rate changes if concentration of A is doubled & that of B is halved ?

A

2

B

4

C

8

D

½

Answer

½

Explanation

Solution

R = k[A]1[B]2

change in rate (2)' (12)2\left( \frac{1}{2} \right)^{2} = 12\frac{1}{2}