Chemistry, 01.07.2019 19:20, elizavlsc4
Co2(g)+ccl4(g)⇌2cocl2(g) calculate δg for this reaction at 25 ∘c under these conditions: pco2pccl4pcocl2===0.140 atm0.185 atm0.735 atm 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.
Answers: 3
for the reaction is 54.425 kJ/mol
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)]](/tpl/images/0039/7802/1c133.png)
![\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)})]](/tpl/images/0039/7802/08d2c.png)
![\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J](/tpl/images/0039/7802/b07a7.png)
for the given reaction:
= Standard gibbs' free energy change of the reaction = 46900 J
![25^oC=[25+273]K=298K](/tpl/images/0039/7802/df1f6.png)
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Co2(g)+ccl4(g)⇌2cocl2(g) calculate δg for this reaction at 25 ∘c under these conditions: pco2pccl4p...
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