Answer:
0.135M/s
Explanation:
We are given that
Initial rate of reaction=0.0300M/s
We have to find the initial rate if [A] is halved and [B] is tripled.
Initial rate,[tex]Rate=k[A][B]^2[/tex]=0.0300M/s
[tex][A]'=\frac{1}{2}[A][/tex]
[tex][B]'=3[B][/tex]
New, initial rate=[tex]k[A]'[B]'^2[/tex]
Substitute the values then we get
New, initial rate,Rate=[tex]k\times \frac{1}{2}[A]\times (3[B])^2[/tex]
New, initial rate
Rate=[tex]4.5k[A][B][/tex]
New initial rate=[tex]4.5\times 0.0300=0.135M/s[/tex]
Hence, the initial rate will be 0.135M/s when [A] is halved and [B] is tripled.
Answer:
The initial rate is 0.135 m/s.
Explanation:
Given that,
Rate = 0.0300 m/s
[tex][A]=\dfrac{[A]}{2}[/tex]
[tex][B]=3[B][/tex]
Suppose, The rate is,
[tex]\text{rate}=k[A][B]^2[/tex]
We need to calculate the initial rate
Using formula of rate
[tex]rate=k[A][B]^2[/tex]
Put the value A and B
[tex]rate=k\times\dfrac{[A]}{2}[3B]^2[/tex]
[tex]rate=k\times\dfrac{[A]}{2}[9B^2][/tex]
[tex]rate=4.5k[A][B]^2[/tex]
[tex]rate =4.5\times0.0300[/tex]
[tex]rate=0.135\ m/s[/tex]
Hence, The initial rate is 0.135 m/s.
