Question

Rubidium, Rb, has a heat of vaporization, Hvap of 69.0 kJ/mol and an entropy of vaporization, Svap of 71.9 J/K mol. Calculate the normal boiling point of rubidium. Group of answer choices

Answer 1

nswer: 959.67 K or 686.67 C

Step-by-step explanation:

Entropy of Vaporization (S) = Heat of Vaporization (H) / T ( boiling point) (in Kelvin)

1.) Flip the equation to have T as the product --> T = H/S

2.) We need to change H from kJ to J --> 69.0 kJ/mol --> 69000J/mol

3.) Substitute numbers in the equation --> (69000 kJ/mol) / (71.9 J/Kmol) = T

4.) T = 959.67 K or in Celcius --> 959.67 - 273 = 686.67 C