Question #a8194

1 Answer
Jan 10, 2017

Answer:

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 = "1 min"#.

At #t = 0#,

#"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 = "1 min"#,

#"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 = "1 min"#,

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

Step 3. Calculate the average rate.

For the reaction

#"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"#

#"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"#