Respuesta :
The answer & explanation for this question is given in the attachment below.



The required activation energy is 0.75 ev/atm.
Given that,
M=26.98 g/mol
T=[tex]500^{\circ}C[/tex](773K)
[tex]N_v=n=7.57\times 10^{23} /m^3[/tex]
Density=2.62 g/[tex]cm^3[/tex]
We know that,
[tex]N=\frac{N_A\rho _{Ad}}{M}\\=\frac{6.02\times10^{23}\times 2.62}{26.98}\\=5.848\times 10^{22} atm/m^3[/tex]
Then,
[tex]N=\frac{N_A\rho _{Ad}}{M}\\=\frac{6.02\times10^{23}\times 2.62}{26.98}\\=5.848\times 10^{22} atm/m^3\\Q_v =-KTln(\frac{N_v}{N})\\=(-8.62\times10^{-5} (773)ln(\frac{7.57\times 10^{23}}{5.843\times10^{28}}\\=0.7499 ev/atm\\\approx 0.75 ev/atm[/tex]
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