# Integrated Rate Equations

1. Zero order reactions:

Order of the reaction is zero.

The reaction in which the rate of reaction is independent of concentration of the reactants.

Rate of reaction remains constant during the course of reaction.

No concentration term in the rate law.

A → Products

Initially t = 0                   a           0

t = t₁                               a – x        x

$$\frac{-d{{C}_{R}}}{dt}=\frac{-d\left[ A \right]}{dt}=\frac{-d\left[ a-x \right]}{dt}=k{{\left( a-x \right)}^{0}}=\frac{dx}{dt}$$

dx/ dt = k

dx = k dt

x = kt + c

When t = 0 x = 0 C= 0

∴ x = kt

t = t½

x = a/2

a/2 = kt½

t½ = (a/2) k

a. Graph for x = kt

b. Graph for t½ = a/2K

c. [A₀] – [A] = kt

[A₀] = initial concentration, t = 0

[A]t = concentration, t = t

[A₀] – [A]t = Kt

+ [A]t = -Kt + [A₀]

Half-life of reaction: The time required for the completion of 50% of the reaction is called half-life of the reaction.

Units of rate constant: n A → Products

Rate law is R = K [Conc]n

K = Rate/ (Concentration)n

= mole¹¯ⁿ literⁿ¯¹ sec¯¹

Where n is order of reaction

2. First order reaction: The reaction in which the rate of R x n depends only on one concentration is doubled. Rate of reaction will also be doubled.

Equations: A → Products

t = 0                   a mole/lit           0

t = t                         a – x                x

Rate = K [A]¹

Rate = – dCR/dt = d [A]/dt = + K [A]

– d (a – x)/ dt = – K (a – x)

dx/ dt = K (a – x)

∫dx/ (a – x) = ∫ Kdt

– log (a – x) = Kt + c

When t = 0, x = 0

– log a = c

– log (a – x) = Kt – log a

Kt = 2030 log (a/a – x)

a = is initial concentration

x = is dissociated concentration.

log (a – x) = – Kt/2.303 + log a

And also

m (a/a-x) = Kt

a/(a – x) = eKt (a – x)/a = e¯Kt

x = a (1 – e¯Kt)

Half-life of first order Reaction:

log (a/a-x) = Kt/2.303

t = 2.303/K log (a/a-x)

t = t½ x = a/2

K = (2.303/ t½) log2 = 0.693/ t½

t½ = 0.693/K [It is independent of initial concentration]

For the first order reaction.

$$M\xrightarrow{1{{t}_{1/2}}}\frac{M}{2}\xrightarrow{2{{t}_{1/2}}}\frac{M}{4}\xrightarrow{3{{t}_{1/2}}}\frac{M}{8}$$

Amount of reaction left after time t = Initial Amount/ 2ⁿ

η = t/t½ = Number of half – lifes

3. nth Order Reaction: A → Products

Rate law is (dn/dt) = K [A]ⁿ = K (a – x)ⁿ

$${{K}_{\eta }}=\frac{1}{\left( n-1 \right)t}\left[ \frac{1}{{{\left( a-x \right)}^{n-1}}}-\frac{1}{{{a}^{n-1}}} \right]$$

Time T is required to complete a particular fraction of reaction.

T α (a)¹¯ⁿ

If concentration is changed m times new rate will be mⁿ

Half-life of nth order reaction:

t½ α 1/aⁿ¯¹

a is initial concentration

η is order of Reaction

Half-life of nth order reaction

$$t\frac{1}{2}=\frac{{{2}^{n-1}}-1}{K\left( n-1 \right){{a}^{n-1}}}$$

$$\frac{{{\left( t\frac{1}{2} \right)}_{1}}}{{{\left( t\frac{1}{2} \right)}_{2}}}={{\left( \frac{{{a}_{2}}}{{{a}_{1}}} \right)}^{n-1}}$$

 Reaction Order Differential Rate Law Integrated Rate Law Characteristics Kinetic Plot Slope of Kinetic Plot Units of Rate Constant Zero – d[A]/dt = K [A] = [A]₀ – Kt [A] vs t – K Mole L¯¹ sec¯¹ First – d[A]/dt = K[A] [A] = [A]₀ e-Kt ln [A] vs t – K sec¯¹ Second – d[A]/dt = K[A]² [A] = [A]₀/1 + Kt[A]₀ 1/[A] vs t K L Mole sec¯¹