Explanation:
The given data is as follows.
T = 298 K, 
= -5645 kJ/mol
= -5798 kJ/mol
Relation between 
and 
are as follows.
= 
-5798 kJ/mol = -5645 kJ/mol - 
-153 kJ/mol = -
= 0.513 kJ/mol K
Now, temperature is 
= (37 + 273) K = 310 K
Since, =
= 
= (-5645 kJ/mol - 159.03 kJ/mol)
= -5804.03 kJ/mol
As, change in Gibb's free energy = maximum non-expansion work
= -5804.03 kJ/mol - (-5798 kJ/mol)
= -6.03 kJ/mol
Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.
Roughness of the surface and weight of the object.
Washing your hands beforehand is important in order to minimize contamination of your experiments. This gets rid of things such as bacteria that can interfere with lab results. You should also wash your hands after completing lab because the chemicals you work with can leave toxic or harmful remains on you. It's important to wash those chemical traces off before you go outside the lab area.
Standard entropy of the compounds of interest are as follows,
= 212.1 J/K.mol
= 205.0 J/K.mol
= 213.6 J/K.mol
= 69.9 J/K.mol
Now for the reaction:
C6H12O6(s) + 6O2(g) —->6CO2(g) + 6H2O(l)
ΔS°reaction = ∑
- ∑
∴ ΔS°reaction =( 6
+ 6
) - (
+ 6
= 205.0 J/K.mol)
∴ ΔS°reaction = [6 (213.6) + 6 (69.9 )] - [(212.1) + 6(205.0)]
∴ ΔS°reaction = 258.9
Answer:
there is no steps shown here
