Answer:
B.Doubling the concentration of B.
Explanation:
Consider the rate below
\(R=k[A][B]^2\)
We have to find the rate of reaction is quadruple means the rate of reaction is 4 times the rate of initial rate of reaction.
Let [A}=a,[B]=b then
\(R_1=kab^2\)
a. doubling the concentration of A
It means [A]=2a
Therefore , substituting the value
Then we get the rate of reaction
\(R_2=k[2a][b]^2=2kab^2=2R_1\)
Therefore , the rate of reaction is 2 times the initial rate of reaction.Hence, option A is false.
B.Doubling the concentration of B
It means [B]=2b
Therefore , substituting the value
Then we get the rate of reaction
\(R_2=k\timesa\times(2b)^2=4 kab^2=4R_1\)
Hence, the rate of reaction is 4 times the initial rate of reaction .Therefore, the rate of reaction is quadruple.Therefore, option B is true.
C.Doubling the concentration of both A and B
[A]=2a, [B]=2b
Substitute the values then the rate of reaction
\(R_2=k(2a)(2b)^2\)
\(R_2=8kab^2=8R_1\)
Hence, the rate of reaction is 8 times the initial rate of reaction .Therefore, it is not quadruple.Therefore, option C is false.
D.Doubling the concentration of A but halving the concentration of B
[A]=2a, [B]=\(\frac{1}{2}b\)
Substitute the values then the rate of reaction
\(R_2=k(2a)(\frac{1}{2}b)^2=\frac{2kab^2}{4}=\frac{1}{2}kab^2\)
\(R_2=\frac{1}{2}R_1\)
Hence, the rate of reaction is \(\frac{1}{2}\) times the rate of initial rate of reaction.Therefore, option D is false.
