Question

The half life of first order reaction is 990s if the initial of the reaction reactant is 0.08 mol dm per cube what concentration after 35 min

Answer 1

k = 0.693 /t 1/2 = 0.693 /990 s = 7 × 10-4 s-1

t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t

t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s,

t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ?

t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10

t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10 [A]0/ [A]t = k t /2.303 = 7 × 10-4 s-1 × 2100 s /2.303 = 0.6383

t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10 [A]0/ [A]t = k t /2.303 = 7 × 10-4 s-1 × 2100 s /2.303 = 0.6383 [A]0/ [A]t = antilog 0.6383 = 4.35

t 1/2 = 0.693 /990 s = 7 × 10-4 s-1 k = 2.303 t log10 [A]0/ [A]t [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, [A]t = ? log10 [A]0/ [A]t = k t /2.303 = 7 × 10-4 s-1 × 2100 s /2.303 = 0.6383 [A]0/ [A]t = antilog 0.6383 = 4.35 Hence, [A]t = [A]0 4.35 = 0.08/ 4.35 = 0.0184 mol dm-3

Answer 2

Given:

Order of reaction = first order

Half-life of reactant = 990s

Initial concentration of the reactant = 0.08 mol/dm³

To find:

The concentration of the reactant after 35min.

Solution:

For a first-order reaction,

$k=\frac{2.303}{t}log\frac{[R_{0}]}{[R]}...(1)$

where $k$ is the rate constant, $[R_{0}]$ is the initial concentration of the reactant and $[R]$ is the final concentration of the reactant and $t$ is the time during which the reaction is completed.

The half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its initial concentration. It is denoted by $t_{1/2}$.

At half-life, $R=\frac{[R_{0}]}{2}$

Hence, the rate constant for a first-order reaction at half-life,

$k=\frac{2.303}{t_{1/2}}log\frac{[R_{0}]}{[R_{0}]/2 }$

∴ $t_{1/2}=\frac{0.693}{k}$

Here, $t_{1/2}=990s$

$k=\frac{0.693}{990}=0.0007$

After a time of 35 minutes, the concentration of the reactant is determined by substituting the values in equation (1),

$k=\frac{2.303}{t}log\frac{[R_{0}]}{[R]}$

$k=0.0007, t=35*60s=2100s, [R_{0}]=0.08 mol/dm^{3}$

$0.0007=\frac{2.303}{2100}log \frac{0.08}{[R]}$

$0.6382=log\frac{0.08}{[R]}$

$e^{0.6382}=\frac{0.08}{[R]}$

$1.89=\frac{0.08}{[R]}$

$[R]=\frac{0.08}{1.89}=0.04mol/dm^{3}$

Hence, the concentration of reactant after 35 mins is 0.04mol/dm³.

The concentration of the reactant in the first-order reaction after 35 mins is 0.04mol/dm³.