Question

Sodium carbonate can be made by heating sodium
bicarbonate:

2NaHCO_{3}(s)-->
Na_{2}CO_{3}(s) + CO_{2}(g) +
H_{2}O(g)

Given that ΔH° = 128.9 kJ/mol and ΔG° = 33.1 kJ/mol at 25°C, What
is the value of ΔS°

Answer in J/mol K to 3 sig figs.

Answer #1

The standard Gibbs free energy ΔG° is related to standard enthalpy change ΔH° and standard entropy change ΔS° as

ΔG° = ΔH°-TΔS°

Where T is the temperature in K

Given that ΔH° = 128.9 kJ/mol

ΔG° = 33.1 kJ/mol

T = 25 C = 273+25 K = 298 K

33.1 kJ/mol = 128.9 kJ/mol - 298K*ΔS°

33.1 kJ/mol -128.9 kJ/mol = - 298K*ΔS°

-95.8 kJ/mol = - 298K*ΔS°

ΔS° = -95.8 kJ/mol/- 298K

= -95.8*1000 J/mol/- 298K

= 321.476 J/mol-K

Rounding off to 3 significant figures we have ΔS° = 321 J/mole-K

Sodium carbonate, Na2CO3, can be made by heating sodium
bicarbonate, NaHCO3, according to the following chemical
equation:
2NaHCO3(s) → Na2CO3(s) + CO2(g) +
H2O(g)
If ΔH° = 128.9 kJ/mol and ΔG° = 33.1 kJ/mol at 25 °C, estimate the
minimum temperature above which
the reaction becomes spontaneous.

Sodium carbonate, Na2CO3(s), can be
prepared by heating sodium bicarbonate, NaHCO3(s).
2NaHCO3(s) ⇌ Na2CO3(s) +
CO2(g) + H2O(g), Kp = 0.264 at 179
°C
If a sample of NaHCO3 is placed in an evacuated flask
and allowed to achieve equilibrium at 179 °C, what will the total
gas pressure (in atm) be?

Consider the reaction:
Ca(s)
+
2 H2O(ℓ)
→
Ca(OH)2(aq)
+
2H2(g)
ΔfH°(kJ/mol)
0
-285.83
-1002.82
0
S°(J/K·mol)
41.59
69.95
-74.5
130.7
What are the values of ΔH°, ΔS° and ΔG°
ΔH° = kJ/mol
ΔS° = J/mol·K
ΔG° = kJ/mol

Sodium carbonate nuetralizes H2SO4 as shown below in
the reaction:
Na2CO3 (s) + H2SO4 (aq) -->Na2SO4 (aq) + H2O (l)
+CO2 (g)
How many grams of sodium carbonate are required to
nuetralize 455ml of 5.4% ( W/V) solution of H2SO4?

Calculate ΔG∘ (in kJ/mol) for the following reaction at 1 atm
and 25 °C:
C2H6 (g) + O2 (g) →
CO2 (g) + H2O (l) (unbalanced)
ΔHf C2H6 (g) = -84.7 kJ/mol; S
C2H6 (g) = 229.5 J/K⋅mol;
ΔHf ∘ CO2 (g) = -393.5 kJ/mol; S
CO2 (g) = 213.6 J/K⋅mol;
ΔHf H2O (l) = -285.8 kJ/mol; SH2O
(l) = 69.9 J/K⋅mol;
SO2 (g) = 205.0 J/K⋅mol

Consider the following reaction: CaCO3(s)→CaO(s)+CO2(g) Estimate
ΔG∘ for this reaction at each of the following temperatures.
(Assume that ΔH∘ and ΔS∘ do not change too much within the given
temperature range.)
Part A
315 K
Express your answer using one decimal place.
ΔG∘ =
kJ
Part B
1075 K
Express your answer using one decimal place.
ΔG∘ =
kJ
Part C
1440 K
Express your answer using one decimal place.
ΔG∘ =
kJ

Consider the following reaction:
CaCO3(s)→CaO(s)+CO2(g).
Estimate ΔG∘ for this reaction at each of the following
temperatures. (Assume that ΔH∘ and ΔS∘ do not
change too much within the given temperature range.)
Part A
310 K
Part B 1035K
Part C 1455K
in kJ

1- Calculate ΔG o for the following reaction at 25°C.
You will have to look up the thermodynamic data. 2 C2H6(g) + 7
O2(g) → 4 CO2(g) + 6 H2O(l)
2- A reaction will be spontaneous only at low
temperatures if both ΔH and ΔS are negative. For
a reaction in which ΔH = −320.1 kJ/mol and ΔS =
−99.00 J/K ·mol,determine the temperature (in °C)below which the
reaction is spontaneous.

Consider the oxidation of CO to CO2:
CO(g)+12O2(g)→CO2(g)
Reactant or product
ΔH∘f(kJ/mol)
S∘(J/mol⋅K)
CO
-110.5
197.7
O2
0
205.2
CO2
-393.5
213.8
Part A
Calculate ΔG∘rxn at 25∘C.
Express your answer to one decimal place with the appropriate
units.
ΔG∘rxn =
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Part B
Determine whether the reaction is spontaneous at standard
conditions.
Determine whether the reaction is spontaneous at standard
conditions.
spontaneous
nonspontaneous

Given the following reactions and their associated enthalpy
changes
CO2 (g) → C (s) +
O2 (g)
ΔH
= 393.5 kJ
C3H8 (g) + 5 O2 (g) → 3 CO2
(g) + 4 H2O (g) ΔH = -2044 kJ
H2 (g) + 1/2 O2 (g)
→ H2O (g)
ΔH
= -241.8 kJ
calculate the enthalpy change for the following reaction:
4 H2 (g) + 3 C (s)→ C3H8 (g)

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