Calculate delta G at 298 K for the following reaction:
CH4(g) + 3Cl2(g) --> CHCl3(g) + 3HCl(g). DeltaGf for CHCl3(g) is -70.4 kJ/mol. (provide answer in kJ)
More info:
HCl deltaH= -92.3 deltaS= 186.9 deltaG= -95.3
CH4 deltaH= -74.6 deltaS= 186.3 deltaG= -50.5
Cl2 deltaH= 0 deltaS= 223.1 deltaG= 0
i need help please! this problem threw me off and i don't know what to do with this CHCl3. details on how you figured out the problem will be awesome. Thank you in advance
CH4(g) + 3Cl2(g) --> CHCl3(g) + 3HCl(g). DeltaGf for CHCl3(g) is -70.4 kJ/mol. (provide answer in kJ)
More info:
HCl deltaH= -92.3 deltaS= 186.9 deltaG= -95.3
CH4 deltaH= -74.6 deltaS= 186.3 deltaG= -50.5
Cl2 deltaH= 0 deltaS= 223.1 deltaG= 0
i need help please! this problem threw me off and i don't know what to do with this CHCl3. details on how you figured out the problem will be awesome. Thank you in advance
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∆G = products - reactants
∆G = (3 x -95.3kJ - 70.4kJ) - (-50.5kJ) = -305.8kJ
∆G = (3 x -95.3kJ - 70.4kJ) - (-50.5kJ) = -305.8kJ