Question

Calculate the standard entropy change for the reaction

2H2(g)+O2(g)?2H2O(l)

using the data from the following table:

Substance	?H?f (kJ/mol)	?G?f (kJ/mol)	S? [J/(K?mol)]
H2(g)	0.00	0.00	130.6
O2(g)	0.00	0.00	205.0
H2O(l)	-285.8	-237.2	69.90

Express your answer to four significant figures and include the appropriate units.

Please show me the steps on how to solve this!!! Thank you!!!

Answer 1

Substance	dHof (kJ/mol)	dGof (kJ/mol)	So J/(K.mol)
H2(g)	0.00	0.00	130.6
O2(g)	0.00	0.00	205.0
H2O(l)	-285.8	-237.2	69.90

1. Pay attention to measurement units kJ or J .        0.001kJ = 1J

2. Remember dS^o = Sum(n_iS^o_product) - Sum(n_jS^o_reactant).

Pay attention to reaction coefficients.

            2H₂(g)   +   O₂(g) =   2H₂O(l)

3. Calculate dS^o_rnx = 2 x 69.90 J/(K.mol) [205 J/(K.mol)+ 2x130.6 J/(K.mol)] =

                        = 139.8 (205 + 261.2) =

                        = 139.8 466.2 =

                        = - 326.4 J/(K.mol)  

dS^o_rnx = - 326.4 J/(K.mol)

with 4 significant figures.

^o means standard condition, _rnx means for reaction, J/(K.mol) means Joule per Kelvin and per mol.

Thats all.

In the same way you can calculate dH^o_rxn or dG^o_rxn.