In a chemical reaction A---->B it is found that rate of reaction doubl...
The given chemical reaction is A ----> B. We are given that the rate of the reaction doubles when the concentration of A is increased four times. From this information, we can determine the order of the reaction with respect to A.
To solve this problem, we need to use the rate equation for the reaction, which is given by the equation:
rate = k[A]^m
where:
- rate is the rate of the reaction
- k is the rate constant
- [A] is the concentration of A
- m is the order of the reaction with respect to A
We are given that when the concentration of A is increased four times, the rate of the reaction doubles. Let's say the initial concentration of A is [A]0 and the rate of the reaction is rate0. When the concentration is increased four times, the new concentration of A is 4[A]0 and the new rate is 2(rate0).
Using the rate equation, we can write:
2(rate0) = k(4[A]0)^m
Simplifying the equation, we get:
2(rate0) = 4^m * k([A]0)^m
Dividing both sides of the equation by k([A]0)^m, we get:
2 = 4^m
Taking the logarithm of both sides, we get:
log(2) = log(4^m)
Using the logarithmic property, we can write:
log(2) = m * log(4)
Simplifying further, we get:
log(2) = 2m * log(2)
Dividing both sides of the equation by log(2), we get:
1 = 2m
Therefore, the order of the reaction with respect to A is 1/2.
Answer: (b) 1/2
In a chemical reaction A---->B it is found that rate of reaction doubl...
It's b... 1/2
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