Here is a reaction: NOBr (g) → NO (g) + $\frac{1}{2}$ Br₂(g)
[NOBr]is measured at different time:
t = 0s ⇒ [NOBr] = 0.0250 M
t = 6.2s ⇒ [NOBr] = 0.0191 M
t = 10.8s ⇒ [NOBr] = 0.0162 M
t = 14.7s ⇒ [NOBr] = 0.0144 M
t = 20.0s ⇒ [NOBr] = 0.0125 M
t = 24.6s ⇒ [NOBr] = 0.0112 M
Calculate the initial half-life of the reaction.

Question

Here is a reaction: NOBr (g) → NO (g) +

\frac{1}{2}

Br₂(g)

[NOBr]is measured at different time:

t = 0s ⇒ [NOBr] = 0.0250 M

t = 6.2s ⇒ [NOBr] = 0.0191 M

t = 10.8s ⇒ [NOBr] = 0.0162 M

t = 14.7s ⇒ [NOBr] = 0.0144 M

t = 20.0s ⇒ [NOBr] = 0.0125 M

t = 24.6s ⇒ [NOBr] = 0.0112 M

Calculate the initial half-life of the reaction.

| General Chemistry 2

Accepted Answer

Determine the order of the reaction:

If the reaction is a first-order reaction:

ln[NOBr] = ln[NOBr]₀ – kt ⇒ we plot ln[NOBr] versus t ⇒ we don’t get a straight line (R² = 0.9848) ⇒ this is not a first-order reaction

If the reaction is a second-order reaction:

1/[NOBr] = 1/[NOBr]₀ + kt
⇒ we plot 1/[NOBr] versus t
⇒ we get a straight line (R² = 1.000) and its equation is y = 2.0028x + 40
⇒ the reaction is a second-order reaction and k = 2.00 M^-1.s^-1

Calculate the initial half-life of the reaction:

for a second-order reaction, t_1/2 =

\frac{1}{k {[A]}_{0}}

=

\frac{1}{2.00 \times 0.0250}

= 20.0s

Integrated rate laws - 1 | Chemical Kinetics: Rate Laws

General Chemistry 2 - Integrated rate laws - 1