The method below can be used to determine the response rate: rate = \(K[A]^x \times[B]^y\)
A is the concentration of reactant A, B I is the concentration of reactant B, x is the order of reaction with respect to A, and y is the order of reaction with respect to reactant B in the above formula.
The rate constant is given as k.
The relationship between the rate and concentration of the reactant is known as the order of reaction.
\(\frac{r_1}{4}\) = \(K[A]^x \times[B]^y\)
\(\frac{r_1}{4\times r_1}=(2)^y\)
\(\frac{1}{4}=(2)^y\)
\(\bigg(\frac{1}{2}\bigg)^2=(2)^y\)
\((2)^{-2}-(2)^y\)
\(\therefore y = -2\)
So, the correct option is (B): -2
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
For $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ $\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following statement is correct?

The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or products per unit time.
Consider the reaction A → B,
Rate of the reaction is given by,
Rate = −d[A]/ dt=+d[B]/ dt
Where, [A] → concentration of reactant A
[B] → concentration of product B
(-) A negative sign indicates a decrease in the concentration of A with time.
(+) A positive sign indicates an increase in the concentration of B with time.
There are certain factors that determine the rate of a reaction: