Question

The moisture content of a gas is often expressed in terms of the dew point. The dew point is the temperature to which the gas must be cooled before the gas becomes saturated with water vapour. At this temperature, water or ice (depending on the temperature) will be deposited on a solid surface. Dew point of $H_{2}O$ is $-{43}^{0}C$ at which vapour pressure of ice formed is $0.07mm$. Assuming that the $CaC{l}_2$ owes its desiccating properties to the formation of $CaC{l}_2\cdot 2H_{2}O$, calculate :
(i) $K_p$ at that temperature of the reaction,
(ii) $\Delta G$
$CaC{l}_2\left(s\right)+2H_{2}O\left(g\right)\rightleftharpoons CaC{l}_2\cdot 2H_{2}O\left(s\right)$

• A. $0.589\times {10}^{8}atm$; $-32.02kJ$
• B. $0.589\times {10}^{8}atm$; $+32.02kJ$
• C. $1.178\times {10}^{8}atm$; $-35.544kJ$
• D. $1.178\times {10}^{8}atm$; $+35.544kJ$

Answer 1

The correct option is C $1.178\times {10}^{8}atm$; $-35.544kJ$

(i) At $-43,P_{H_{2}O}=0.07mm=\frac{0.07}{760}atm$
For, $CaC{l}_2\left(s\right)+2H_{2}O\left(g\right)\rightleftharpoons CaC{l}_2\cdot 2H_{2}O\left(s\right)$
$K_p=\frac{1}{{\left(P_{H_{2}O}\right)}^2}=\frac{1}{{\left(\frac{0.07}{760}\right)}^2}$
$=1.178\times {10}^{8}atm$
(ii) $\Delta G=-2.303RT\log K_p(\because T=273-43=230K)$
$=-2.303\times 8.314\times 230\times \log [1.178\times {10}^8]$
$=-35544.05J$
$=-35.544kJ$