Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.

This question is incomplete, the complete question is;

Tonksite is a solid at 300.00K.  At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.

the sublimation pressure of the solid at the melting point is  0.3727 atm

Explanation:

Given that;

T1 = 300 K

T2 = 400 K

H_sub = 66 kJ/mol = 66000 J/mol

P1 = 5.00 × 10⁻⁴ atm

p2 = ?

now using the expression

log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R ×  2.303 ) (( T2 - T1) / T1T2)

now we substitute of given values into the expression

log(p2/p1) = (66000 / 8.314 ×  2.303 ) (( 400 - 300) / 300 × 400 )

p2 = 0.3727 atm

therefore the sublimation pressure of the solid at the melting point is  0.3727 atm

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.