Question

Calculate the change in enthalpy of 1.94 mol of PbO(s) if it is cooled from 732 K to 234 K at constant pressure.

Answer 1

Complete question:

The heat capacity of solid lead oxide is given by Cp,m=44.35+1.47×10⁻³T/K in units of J K−1 mol−1.

Calculate the change in enthalpy of 1.94 mol of PbO(s) if it is cooled from 732 K to 234 K at constant pressure.

Answer:

The change in enthalpy of PbO(s) is -39.488 x10³J

Explanation:

Given:

Initial temperature of PbO(s) (T₁) = 732 K

Final temperature of PbO(s) (T₂) = 234 K

$\delta H = n\int\limits^{T_2}_{T_1} {C_p m} \, dT$

where;

Cp,m is the specific heat capacity of PbO(s)

$\delta H = 1.94 molX\int\limits^{234}_{732} {[44.35 +1.47 X10^{-3}\frac{T}{K} ]} \,d (\frac{T}{K})$

$\delta H = 1.94 molX {[44.35 (234-732) +1.47 X10^{-3}(\frac{234^2 -732^2}{2}) ]}$

     = 1.94mol [(-19957.5)+(-396.9)]

     = -38717.55 J -769.986J

     = -39487.536 J

ΔH = -39.488 x10³J

Therefore, the change in enthalpy of PbO(s) is -39.488 x10³J