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Question: The decomposition of \[N{}_2O{}_5\]dissolved in carbon tetrachloride occurs as follows. \[N{}_2O{}...

The decomposition of N2O5N{}_2O{}_5dissolved in carbon tetrachloride occurs as follows.
N2O5(solution)⇌2NO2(solution)+12O2(g)N{}_2O{}_{5(solution)} \rightleftharpoons 2NO{}_{2(solution)} + \dfrac{1}{2}O{}_{2(g)}
This reaction is first order and its rate constant is 5.0×10−4sec⁡−15.0 \times 10{}^{ - 4}\sec {}^{ - 1}. If initial concentration of N2O5N{}_2O{}_5 for this reaction is 0.5×10−40.5 \times 10{}^{ - 4}molelitre−1litre{}^{ - 1}, then
A. What will be the initial reaction rate?
B. What will be the half life period of this reaction?
C. What will be the concentration of N2O5N{}_2O{}_5 and NO2NO{}_2 at the end of 5050 minutes after the starting of the reaction?

Explanation

Solution

To solve the following question we need to know the chemical kinetics of the given reaction. Chemical kinetics is contrasted with thermodynamics, which deals with the direction in which the process will occur. It investigates experimental conditions that influence the speed of chemical reaction and gives the information about the reactions transition as well as mechanism states which help us describe the characteristics of a chemical reaction.

Complete step by step answer:
A reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance. It is given by the expression
Rate=−d[N2O5]dt=k[N2O5]1 = \dfrac{{ - d[N{}_2O{}_5]}}{{dt}} = k[N{}_2O{}_5]{}^1
In the following question,
k=5×10−4s−1k = 5 \times 10{}^{ - 4}s{}^{ - 1}and [N2O5]=0.5M[N{}_2O{}_5] = 0.5M
When these values are substituted we get,
Initial rate
=5×10−4×0.5 =2.5×10−4M/s  = 5 \times 10{}^{ - 4} \times 0.5 \\\ = 2.5 \times 10{}^{ - 4}M/s \\\
B) The half life of a reaction is defined as the time taken for the concentration of a reactant to reach half of its initial concentration,i.e. the time taken for a reactant concentration to reach half of its initial reaction. It denoted by t12t{}_{\dfrac{1}{2}}
Half life for first order reaction is given as ;
=t12,[N2O5]=2[N2O5]02 =t12=0.69k =t12=1380s =23min⁡  = t{}_{\dfrac{1}{2}},[N{}_2O{}_5] = \dfrac{{2[N{}_2O{}_5]{}_0}}{2} \\\ = t{}_{\dfrac{1}{2}} = \dfrac{{0.69}}{k} \\\ = t{}_{\dfrac{1}{2}} = 1380s \\\ = 23\min \\\
C) The rate law for a chemical reaction is the reaction rate with concentration of the reactants. For the general reaction aA+bB→CaA+bB→CaA + bB \to CaA + bB \to Cwhere there is no intermediate steps in its reaction mechanism. This law is given by r=k[A]x[B]yr = k[A]{}_x[B]{}_y
Rate law of first order reaction is given as,
ln⁡([N2O5]0[N2O5]t)=kt\ln \left( {\dfrac{{[N{}_2O{}_5]{}_0}}{{[N{}_2O{}_5]{}_t}}} \right) = kt
=ln⁡(0.5x)=5×10−4×3000 =x=[N2O5]=0.11M =[NO2]=2×(0.50−0.11) =0.78M  = \ln \left( {\dfrac{{0.5}}{x}} \right) = 5 \times 10{}^{ - 4} \times 3000 \\\ = x = [N{}_2O{}_5] = 0.11M \\\ = [NO{}_2] = 2 \times (0.50 - 0.11) \\\ = 0.78M \\\
Note: We should keep in mind that there are lots of factors that affect the rate of reaction such as the nature of the reactant, physical state, and concentration, the surface area of solid state, temperature, catalysts, pressure and absorption of light.