Question

What is the approximate thermal energy in kJ/mol of molecules at 275° C? Express your answer to the nearest 0.1 kJ/mol

Answer 1

Its thermal energy for 1 molecule is given by

∆E = 3/2 KT

where K = boltzmann constant = 1.3807 x 10 ^-23 J. K ^-1

here temperature is in kelvin

so we have to convert first degree celsius to kelvin

275°C = 275 +273.15 K = 548.15 K

Thermal energy for 1 molecule

∆E = 3/2 *1.3807 x 10 ^-23 J. K ^-1 *548.15 K = 1.135 x 10 ^-20 J/molecule

for 1 mole = 6.023x 10 ²³ molecules/mol

  ∆E_{1 mole} = 1.135 x 10 ^-20 J/molecule/mol x 6.023x 10 ²³ molecules = 6837.58 J/mol

or 6.8 kJ as 1000 J = 1kJ

Ans = 6.8 kJ/mol