Chemistry: Consider the following reaction: co2(g)+ccl4(g)⇌2cocl2(g) calculate δg for this reaction at25 ∘c under these conditions: pco2pccl4pcocl2===0.100atm0.180atm0.735atm δg∘f for co2(g) is −394.4kj/mol, δg∘f for ccl4(g) is −62.3kj/mol, and δg∘f for cocl2(g) is −204.9kj/mol.

Explanation:Just trying to get some points...

Consider the following reaction: co2(g)+ccl4(g)⇌2cocl2(g)

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30.12.2021

The $\Delta G$ for the reaction is 55.328 kJ/mol

Explanation:

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:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.100atm\\p_{CCl_4}=0.180atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.100\times 0.180}\\\\K_p=30.0125$

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 = 30.0125

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(30.0125))\\\\\Delta G=55327.74J/mol=55.328kJ/mol$

Hence, the $\Delta G$ for the reaction is 55.328 kJ/mol

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