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Question: The decomposition of\({{N}_{2}}{{O}_{5}}\) according to the equation \(2{{N}_{2}}{{O}_{5}}(g)\to 4N{...

The decomposition ofN2O5{{N}_{2}}{{O}_{5}} according to the equation 2N2O5(g)→4NO2(g)+O2(g)2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g) is a first – order reaction. After 30 minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be 284.5 mm of Hg and on complete decomposition; the total pressure is 584.5 mm of Hg.
The rate constant of the reaction is?

Explanation

Solution

A first order reaction is the reaction where the concentration of the reacting species is raised to the power of one. The rate constant of any reaction is the constant of proportionality that is introduced in the rate law, that rate of any reaction is directly proportional to the concentration of reactants raised to the stoichiometric coefficients.

Complete answer:
We have been given the decomposition of a gas N2O5{{N}_{2}}{{O}_{5}} that follows the first order reaction. The concentration is given in units of pressure, where after 30 minutes the pressure becomes 284.5 mm of Hg and on complete decomposition the total pressure is 584.5 mm of Hg. We have to find the rate constant.
So, first giving the concentration, variables according to the reaction as:
                                     2N2O5(g)→4NO2(g)+O2(g) initial pressure             a                           0                  0 at time t                           a−x                    2x At complete                   0                           2a dissociation \begin{aligned} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g) \\\ & initial\,pressure\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\\ & at\,time\,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\\ & At\,complete\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a \\\ & dissociation \\\ \end{aligned}
So, total number of moles at time, t = a−x+2x+x2=a+32xa-x+2x+\dfrac{x}{2}=a+\dfrac{3}{2}x
After complete dissociation the total moles will be = 2a+a2=52a2a+\dfrac{a}{2}=\dfrac{5}{2}a
As, we know the number of moles are directly proportional to the pressure, so,
52a∝584.5mm Hg....(1)\dfrac{5}{2}a\propto 584.5mm\,Hg....(1)
a+32x∝284.5mm Hg.......(2)a+\dfrac{3}{2}x\propto 284.5mm\,Hg.......(2)
After solving equations 1 and 2 we have, a = 233.8 mm Hg and x = 33.8 mm Hg. Substituting the values of a and x (concentration at time t and final concentration after dissociation) in the first order equation we have,
K=2.303tlog⁡aa−xK=\dfrac{2.303}{t}\log \dfrac{a}{a-x}
K=2.30330 min⁡log⁡233.8233.8−33.8K=\dfrac{2.303}{30\,\min }\log \dfrac{233.8}{233.8-33.8}
K = 5.21×10−3min⁡−15.21\times {{10}^{-3}}{{\min }^{-1}}
So, the rate constant of the given reaction is 5.21×10−3min⁡−15.21\times {{10}^{-3}}{{\min }^{-1}}.

Note:
The integrated form of the rate equation is used here that has rate constant equal to the negative log by time into the final concentration after decomposition upon initial concentration at time t. For graphical representation of concentration versus time, the negative slope of the graph gives us the rate constant.