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Question
The decomposition of N2O5 in CCl4 at 318 K has been studied by monitoring the concentration of N2O5 in the solution. Initially the concentration of N2O5 is 2.33 (mol L−1 and after 184 minutes, it is reduced to 2.08 mol L−1. The reaction takes place according to the equation
2N2O5(g)→4NO2(g)+O2(g)
Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO2 during this period?
2N2O5(g)→4NO2(g)+O2(g)
Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO2 during this period?
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Solution
Given,
Initial concentration = 2.33 mol L−1
Final concentration = 2.08 mol L−1
Time interval = 184 minutes
Average rate
Average rate of the given reaction can be written as-
Average rate= 12{−Δ[N2O5]Δt}14{−Δ[NO2]Δt}
=−12[(2.08−2.33)mol L−1184 min]
=−12[(2.08−2.33)mol L−1184 min]
=6.79×10−4mol L−1min
=(6.79×10−4mol L−1min−1)×(60 min/1h)
=4.07×10−2mol L−1/h
=(6.79×10−4mol L−1×1min/60s)
=1.13×10−5mol L−1s−1
Rate of formation of NO2
Rate of formation of NO2 can be written as -
{Δ[NO2]Δt}=4×Average rate
{Δ[NO2]Δt}=6.79×10−4×4mol L−1min−1
=2.72×10−3mol L−1min−1
Final answer:
1. Average rate of reaction
=6.79×10−4mol L−1/min
=4.07×10−2mol L−1/h
=1.13×10−5mol L−1s−1
2. Rate of production of NO2
=2.72×10−3mol L−1min−1.
Initial concentration = 2.33 mol L−1
Final concentration = 2.08 mol L−1
Time interval = 184 minutes
Average rate
Average rate of the given reaction can be written as-
Average rate= 12{−Δ[N2O5]Δt}14{−Δ[NO2]Δt}
=−12[(2.08−2.33)mol L−1184 min]
=−12[(2.08−2.33)mol L−1184 min]
=6.79×10−4mol L−1min
=(6.79×10−4mol L−1min−1)×(60 min/1h)
=4.07×10−2mol L−1/h
=(6.79×10−4mol L−1×1min/60s)
=1.13×10−5mol L−1s−1
Rate of formation of NO2
Rate of formation of NO2 can be written as -
{Δ[NO2]Δt}=4×Average rate
{Δ[NO2]Δt}=6.79×10−4×4mol L−1min−1
=2.72×10−3mol L−1min−1
Final answer:
1. Average rate of reaction
=6.79×10−4mol L−1/min
=4.07×10−2mol L−1/h
=1.13×10−5mol L−1s−1
2. Rate of production of NO2
=2.72×10−3mol L−1min−1.
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