Respuesta :
Answer: The time taken for the concentration of nitrogen dioxide to decrease is 54.0 seconds
Explanation:
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
[tex]\ln(\frac{K_{479K}}{K_{273K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_{479K}[/tex] = equilibrium constant at 479 K = ?
[tex]K_{273K}[/tex] = equilibrium constant at 273 K = [tex]2.3\times 10^{-12}L.mol^{-1}s^{-1}[/tex]
[tex]E_a[/tex] = Activation energy = 111 kJ/mol = 111000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = 273 K
[tex]T_2[/tex] = final temperature = 479 K
Putting values in above equation, we get:
[tex]\ln(\frac{K_{479K}}{2.3\times 10^{-12}})=\frac{111000J}{8.314J/mol.K}[\frac{1}{273}-\frac{1}{479}]\\\\K_{479K}=3.132\times 10^{-3}L.mol^{-1}s^{-1}[/tex]
The given chemical equation follows:
[tex]2NO_2(g)\rightarrow 2NO(g)+O_2(g)[/tex]
As, the above equation is bimolecular elementary reaction, the order od the reaction is 2.
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{p}=kt+\frac{1}{p_o}[/tex]
[tex]p_o[/tex] = initial partial pressure = 2.80 atm
p = partial pressure left after time t = 1.90 atm
k = rate constant = [tex]3.132\times 10^{-3}L.mol^{-1}s^{-1}[/tex]
t = time taken = ?
Putting values in above equation, we get:
[tex]\frac{1}{1.90}=(3.132\times 10^{-3}\times t)+\frac{1}{2.80}[/tex]
[tex]t=54.0s[/tex]
Hence, the time taken for the concentration of nitrogen dioxide to decrease is 54.0 seconds
