# Quiz | Chemical Kinetics: Rate Laws

What is the rate of the following reaction: 2A + B → C?

$-\frac{\u2206\left[\mathrm{A}\right]}{\u2206\mathrm{t}}$

$\frac{\u2206\left[\mathrm{A}\right]}{\u2206\mathrm{t}}$

$-\frac{1}{2}\frac{\u2206\left[\mathrm{A}\right]}{\u2206\mathrm{t}}$

$\frac{1}{2}\frac{\u2206\left[\mathrm{A}\right]}{\u2206\mathrm{t}}$

A is a reagent and its stochiometric coefficient is 2. Reaction rate of a reagent:

- $\frac{1}{\mathrm{\alpha}}\frac{\u2206\left[\mathrm{reagent}\right]}{\u2206\mathrm{t}}$

α = stochiometric coefficient

[reagent] = concentration of the reagent (in mol.L^{-1})

What is the overall order of a reaction with a rate = k [A]^{2 }[B]?

0

1

2

3

The overall order of reaction is the sum of the orders of reaction of all the reactants

What are the units of the constant k in the rate of reaction = k [A]^{2 }[B] (concentration in mol.L^{-1}) ?

s

mol^{-2}.L^{2}.s

mol^{-1}.L^{1}.s^{-1}

mol^{-2}.L^{2}.s^{-1}

The rate of reaction is in mol.L^{-1}.s^{-1} and [A]^{2 }[B] is in mol^{3}.L^{-3} ⇒ k is in mol^{-2}.L^{2}.s^{-1}

What is correct concerning the half-time of a first-order reaction?

t_{1/2} is independent of the initial concentration of the reactant

t_{1/2} is proportional to the initial concentration of the reactant

t_{1/2} is inversely proportional to the initial concentration of the reactant

None of the above

For a first-order reaction: ${\mathrm{t}}_{1/2}=\frac{\mathrm{ln}2}{\mathrm{k}}$

What is correct concerning the half-time of a second-order reaction?

t_{1/2} is independent of the initial concentration of the reactant

t_{1/2} is proportional to the initial concentration of the reactant

t_{1/2} is inversely proportional to the initial concentration of the reactant

None of the above

For a second-order reaction: ${\mathrm{t}}_{1/2}=\frac{1}{\mathrm{k}{\left[\mathrm{A}\right]}_{0}}$

What is the relationship between [A] and t for a first-order reaction?

ln[A] = ln[A]_{0} + kt

$\frac{\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}_{0}}$ = e^{-kt}

$\frac{\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}_{0}}$ = e^{t}

$\frac{1}{\left[\mathrm{A}\right]}$ = $\frac{1}{{\left[\mathrm{A}\right]}_{0}}$ + kt

For a first-order reaction: $-\frac{\u2206\left[\mathrm{A}\right]}{\u2206\mathrm{t}}$ = k [A]

After integrating: ln [A] = ln [A]_{0} - kt ⇒ $\frac{\left[\mathrm{A}\right]}{{\left[\mathrm{A}\right]}_{0}}$ = e^{-kt}