Question

The reaction between X and Y is first order with respect to X and second order with respect to Y if concentration of X is halved and concentration of Y is doubled the rate of reaction will be​

Answer 1

Explanation:

If a reactant is first order, then when its concentration is doubled, the rate of the reaction doubles; if the concentration is tripled, the rate triples, etc. (If a reactant is third order, then when its concentration is doubled, the rate of reaction increases by a factor of 8 (23 = 8), etc.)

Answer 2

Answer: The rate of the reaction will become double the initial value.

Explanation:

According to the rate law:

$r(rate)= K [X]^{a} [Y]^{b}$

Where, a and b are the order of reactants X and Y.

$a=1\\b=2$

then,

$r(rate)= K [X]^{1} [Y]^{2}$

Now the concentration of X and Y is given half and double the initial value.

Then, the rate of the reaction will be:

$r'(rate)= K [\frac{X}{2} ]^{1} [2Y ]^{2}\\r'=K \frac{[X]^{1}[Y]^{2} }{2}$

We can write:

$r'=\frac{r}{2}$

Hence, the correct answer is the rate will become half the initial value.

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