# Question a8194

Jan 10, 2017

The average rate of reaction is  1.82 × 10^"-4" color(white)(l)"mol·L"^"-1""s"^"-1".

#### Explanation:

Step 1. Calculate the moles at $t = 0$ and $t = \text{1 min}$.

At $t = 0$,

${\text{Moles of NO"_2 = 2.179 color(red)(cancel(color(black)("g NO"_2))) × ("1 mol NO"_2)/(46.01 color(red)(cancel(color(black)("g NO"_2)))) = "0.047 359 mol NO}}_{2}$

At $t = \text{1 min}$,

${\text{Moles of NO"_2 = 1.879 color(red)(cancel(color(black)("g NO"_2))) × ("1 mol NO"_2)/(46.01 color(red)(cancel(color(black)("g NO"_2)))) = "0.040 839 mol NO}}_{2}$

Step 2. Calculate the concentrations at $t = 0$ and $t = 1 \min$.

At $t = 0$,

["NO"_2] = "0.047 359 mol"/"0.300 L" = "0.1579 mol/L"

At $t = \text{1 min}$,

["NO"_2] = "0.040 839 mol"/"0.300 L "= "0.1361 mol/L"

Step 3. Calculate the average rate.

For the reaction

${\text{2NO"_2 → "2NO" + "O}}_{2}$

the rate of reaction is defined as

"rate" = -1/2(Δ["NO"_2])/(Δt)

(Δ["NO"_2])/(Δt) = (["NO"_2]_2 - ["NO"_2]_1)/(t_2 - t_1) = "0.1361 mol/L - 0.1579 mol/L"/"1 min - 0" = "-0.0218 mol·L"^"-1""min"^"-1"

$\text{rate" = -1/2("-0.0218 mol·L"^"-1""min"^"-1") = "0.0109 mol·L"^"-1"color(red)(cancel(color(black)("min"^"-1"))) × (1 color(red)(cancel(color(black)("min"))))/"60 s}$
= 1.82 × 10^"-4" color(white)(l)"mol·L"^"-1""s"^"-1"#