Answer: Option (d) is the correct answer.
Explanation:
Rate law is the rate which is in equilibrium with the concentration of reactants raised to the power of their coefficients.
For example, \(xA + yB \rightarrow zC\)
Rate, r = k \([A]^{x} \times [B]^{y}\) ........ (1)
So, when concentration of A is doubled then rate of reaction also doubles. Hence, then rate law will be as follows.
2r = k \([2A]^{x} \times [B]^{y}\) ......... (2)
Dividing equation (1) by (2) we get the following.
\(\frac{r}{2r}\) = \(\frac{k[A]^{x} \times [B]^{y}}{k[2A]^{x} \times [B]^{y}}\)
Therefore, cancelling common factors we get the following.
\(\frac{1}{2} = \frac{1}{2x}\)
\(2^{x}\) = 2
x = 1
Thus, we can conclude that in the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not affect the reaction rate then rate of reaction will be as follows.
rate = k[A]
