Question

Calculate delta G rxn and delta S theoretical for the reaction. Using delta G rxn and the average of the experimentally determined value for delta H in experiment, report a semi-experimental value for delta S (assume standard temp). Compare value to the theoretical value and calculate percent error. Does the sign on the experimental value agree with the type of change?

Component | delta G (kJ/mol) | S (J/mol x K) |

Ca (s) | 0 | 41.6 |

H20 (l) | -237.13 | 69.9 |

Ca(OH)2(s) | -897.5 | 83.4 |

H2(g) | 0 | 131 |

Found average of the experimental delta H to be 3.29 x
10^{5}

I keep finding my percent error to be extremely drastic. I don't know if I am doing the steps wrong, or if my experimental delta H is wrong. Appreciate the help.

Answer #1

We have reaction:

Ca(s) + 2H_{2}O(l) ------>Ca(OH)_{2}(s) +
H_{2}(g)

overall ∆Grxn = ∆G of products - ∆G of reactants

-897.5 kJ/mol x 1 mol + 0 x 1 mol -(2 mol x -237.13 kJ/mol + 0 x 1mol)

= -423.24 kJ = -423240 J

∆S = ∆S of products - ∆s of reactants = 83.4 J/mol.K x 1mol+ 131J/mol.K x 1mol - (2 mol x 69.9 J/mol.K + 41.6 J/mol.K x 1 mol)

= 34 J/.K

∆G = ∆H -T∆S

**Standard temperature** is defined as zero degrees
Celsius (0 ^{0}C) or 273.15 K.

-423240 J = ∆H - 273.15 K x 34 J/mol.K

∆H = -413952.9 J = -4.139 x 10^{5} J

% error = {-4.139 x 10^{5} J -(-3.29 x
10^{5}J)}/-4.139 x 10^{5} J x 100 = 20.52 %

you might have forgot to convert the units from kJ to J

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propane combustion, using thermochemical equations below. Show all
work! 3 C (s) + 4 H2 (g) = C3H8 (g) Delta H = (-) 103.9 kJ/mol C
(s) + O2 (g) = CO2 (g) Delta H = (-) 393.5 kJ/mol H2 (g) + 1/2 O2
(g) = H2O (g) Delta H = (-)241. 8 kJ/mol Delta H rxn : ?

Given that: 2Al(s) + 3/2O2(g) -->
Al2O3(s) Delta H degrees rxn =-1601kJ/mol
and 2Fe(s) + 3/2O2(g) -->
Fe2O3(s) Delta H degrees rxn =-821kJ/mol
Calculate the standard enthalpy change for the following
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Al2O3(s)
Answer in scientific notation

Use the table to calculate delta G rxn for the following
reaction at 25 degree C.
A.) 2Mg(s) +O2(g) --> 2MgO(s)
B.) 2SO2(g) +O2(g) --> 2SO3(g)
C.) 2C2H2(g) + 5O2(g) --> 4CO2(g) + 2H2O(l)
Species
Delta G (kJ/mol)
Mg(s) 0
O2
(g) 0
MgO(s) -255.0
SO2(g)
-300.57
SO3(g)
-370.53
C2H2(g)
168.43
CO2(g)
-394.65
H2O(l)
-237.35
H2O(g)
-228.77

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(d) Calculate the thermodynamic equilibrium constant for this
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Find H for the reaction 2 ZnS(s) + 3 O2(g) ---> 2 ZnO(s) +
2SO2(g)
delta H f (ZnS,s) = -206.0 Kj/mol
delta H f (ZnO,s) -348.3 Kj/mol
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N2 delta H=0.00kj mol^-1s=+191.5J mol^-1K^-1
H2 delta H=0.00kj mol^-1,s = +130.6j mol^-1 k-1
NH3 delta H=-46.0kj mol^-1,s =192.5 J mol^-1 k-1
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B.-87.6kJ
C.-7.4kJ
D.-32.9 kJ
E.-151.1kJ

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2Ni(s) + 2S(s) +3O2 (g) -> 2NiSO3(s)
from the following info:
NiSO3(s) -> NiO(s) + SO2(g) delta H=156kJ
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