Question

Temperature at which the equilibrium water vapour pressure is 1.00 bar.

• A. ${107}^0$ C
• B. ${380}^0$ C
• C. ${215}^0$ C
• D. ${240}^0$ C

Answer 1

The correct option is A ${107}^0$ C

$K=P_{H_{2}O}^{1.5}=1^{1.5}=1$

$\Delta H^0=H_{CaS{O}_4.0.5H_{2}O}^0+1.5H_{H_{2}O}^0-H_{CaS{O}_4.2H_{2}O}^0$

$\Delta H^0=-1575.0-1.5\times 241.8+2021.0=83.3kJ/mol$

$\Delta S^0=S_{CaS{O}_4.0.5H_{2}O}^0+1.5S_{H_{2}O}^0-S_{CaS{O}_4.2H_{2}O}^0$

$\Delta S^0=130.5+1.5\times 188.6-194.0=219.4J/K.mol$

Let T be the equilibrium temperature.

${\Delta }^{0}G={\Delta }^{0}H-T{\Delta }^{0}S$

${\Delta }^{0}G=83300-219.4T$....(1)

Also,${\Delta }^{0}G=-RTlnK$

${\Delta }^{0}G=-8.314\times T\times ln1=0$......(2)

From (1) and (2),

$83300-219.4T=0$

$T=380K=380-{273}^{0}C={107}^{0}C$