The normal boiling point of ammonia is -33.34 $\degree$C, and its enthalpy of vaporization is 23.35 kJ/mol.What pressure would have to be applied for ammonia to boil at 25.00 $\degree$C?

Question

Quizplus · Accepted Answer

The Clausius-Clapeyron equation is:ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)where P1 is the pressure at temperature T1, P2 is the pressure at temperature T2, ΔHvap is the enthalpy of vaporization, and R is the gas constant.We can rearrange this equation to solve for P2:P2 = P1 * exp[(ΔHvap/R) * (1/T1 - 1/T2)]Plugging in the given values, we get:P2 = 1 atm * exp[(23.35 kJ/mol / 8.314 J/mol-K) * (1/298.15 K - 1/298.15 K + 25 K)]P2 = 1 atm * exp(0.0087)P2 = 1.009 atmTherefore, the pressure that would have to be applied for ammonia to boil at 25.00°C is 1.009 atm.

The Normal Boiling Point of Ammonia Is -33 °\degree° C, and Its Enthalpy of Vaporization Is 23

The Normal Boiling Point of Ammonia Is -33 $\degree$ C, and Its Enthalpy of Vaporization Is 23