In a reaction, $A + B \rightarrow \text{Product}$, the rate is doubled when the concentration of $B$ is doubled, and the rate increases by a factor of $8$ when the concentrations of both the reactants ($A$ and $B$) are doubled. The rate law for the reaction can be written as:

If the half-life is independent of its initial concentration, then the order of the reaction is:

Which quantity is altered when a catalyst is introduced during a chemical reaction?

The rate of the reaction $2NO + Cl_2 \rightarrow 2NOCl$ is given by the rate equation $\text{rate} = k[NO]^2[Cl_2]$. The value of the rate constant can be increased by:

The rate of a first-order reaction is $1.5 \times 10^{-2} \text{ mol L}^{-1} \text{ min}^{-1}$ at $0.5 \text{ M}$ concentration of the reactant. The half-life of the reaction is:

The activation energy of a reaction can be determined from the slope of the graph between:

The plot of $\ln k$ vs $1/T$ for the following reaction, $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ gives a straight line with the slope of the line equal to $-1.0 \times 10^4 \text{ K}$. What is the activation energy for the reaction in $\text{J mol}^{–1}$? (Given: $R = 8.3 \text{ J K}^{–1} \text{ mol}^{–1}$)

Consider the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$. The equality relationship between $\frac{d[NH_3]}{dt}$ and $-\frac{d[H_2]}{dt}$ is:

For a chemical reaction, $4A + 3B \rightarrow 6C + 9D$ rate of formation of C is $6 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$ and rate of disappearance of A is $4 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}$. The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be: