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Allisa [31]
3 years ago
11

What effect has impurities on the melting and boiling points of pure substances? illustrate your answer with an example

Chemistry
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

The presence of an impurity in a substance lowers the melting point, and raises the boiling point of the substance. For example, Impurities present in the water decrease the freezing temperature of the water, and in the same way, impurities increase the boiling point of the water.

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

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