Question

The chemical reaction that causes iron to corrode in air is given by

4Fe+3O2→2Fe2O3

in which at 298 K

ΔH∘rxn |
= −1684 kJ |

ΔS∘rxn |
= −543.7 J/K |

**Part A:**

What is the standard Gibbs free energy for this reaction? Assume the commonly used standard reference temperature of 298 K.

**Part B:**

What is the Gibbs free energy for this reaction at 3652 K ?
Assume that Δ*H* and Δ*S* do not change with
temperature.

**Part C:**

At what temperature *T*eq do the forward and reverse
corrosion reactions occur in equilibrium?

Answer #1

A)

ΔH = -1684.0 KJ

ΔS = -543.7 J/K

= -0.5437 KJ/K

T = 298 K

use:

ΔG = ΔH - T*ΔS

ΔG = -1684.0 - 298.0 * -0.5437

ΔG = -1521.9774 KJ

Answer: -1522 KJ

B)

Now we have:

ΔH = -1684.0 KJ

ΔS = -543.7 J/K

= -0.5437 KJ/K

T = 3652 K

use:

ΔG = ΔH - T*ΔS

ΔG = -1684.0 - 3652.0 * -0.5437

ΔG = 302 KJ

Answer: 302 KJ

C)

ΔG = 0.0 KJ (because this is equilibrium)

ΔH = -1684.0 KJ

ΔS = -543.7 J/K

= -0.5437 KJ/K

use:

ΔG = ΔH - T*ΔS

0.0 = -1684.0 - T *-0.5437

T = 3097 K

Answer: 3097 K

The chemical reaction that causes iron to corrode in air is
given by
4Fe+3O2→2Fe2O3
in which at 298 K
ΔH∘rxn
= −1684 kJ
ΔS∘rxn
= −543.7 J/K
Gibbs free energy (G) is a measure of the spontaneity of a
chemical reaction. It is the chemical potential for a reaction, and
is minimized at equilibrium. It is defined as G=H−TS where H is
enthalpy, T is temperature, and S is entropy.
Part A
What is the standard Gibbs free energy for...

The chemical reaction that causes iron to corrode in air is
given by 4Fe(s)+3O2(g)→2Fe2O3(s)
and
ΔrH∘ = −1684 kJ mol−1
ΔrS∘ = −543.7 J K−1 mol−1
a) What is the standard Gibbs energy change for this reaction?
Assume the commonly used standard reference temperature of 298
K.
b) What is the Gibbs energy for this reaction at 3652 K ? Assume
that ΔrH∘ and ΔrS∘ do not change with
temperature.
c) The standard Gibbs energy change, ΔrG∘, applies only
when...

A. Using given data, calculate the change in
Gibbs free energy for each of the following reactions. In each case
indicate whether the reaction is spontaneous at 298K under standard
conditions.
2H2O2(l)→2H2O(l)+O2(g)
Gibbs free energy for H2O2(l) is -120.4kJ/mol
Gibbs free energy for H2O(l) is -237.13kJ/mol
B. A certain reaction has ΔH∘ = + 35.4
kJ and ΔS∘ = 85.0 J/K . Calculate ΔG∘ for the
reaction at 298 K. Is the reaction spontaneous at
298K under standard
conditions?

he thermodynamic properties for a reaction are related by the
equation that defines the standard free energy, ΔG∘, in
kJ/mol:
ΔG∘=ΔH∘−TΔS∘
where ΔH∘ is the standard enthalpy change in kJ/mol and
ΔS∘ is the standard entropy change in J/(mol⋅K). A good
approximation of the free energy change at other temperatures,
ΔGT, can also be obtained by utilizing this
equation and assuming enthalpy (ΔH∘) and entropy
(ΔS∘) change little with temperature.
Part A
For the reaction of oxygen and nitrogen to...

For a given reaction, ΔH = -26.6 kJ/mol and ΔS = -77.0 J/K⋅mol.
The reaction is spontaneous ________. Assume that ΔH and ΔS do not
vary with temperature.
For a given reaction, = -26.6 kJ/mol and =
-77.0 J/Kmol. The reaction is spontaneous ________. Assume
that and do not vary with temperature.
at T > 298 K
at all temperatures
at T < 345 K
at T < 298 K
at T > 345 K

Assume we know the Gibbs free energy of reaction for a chemical
reaction taking place at 298 K. What additional
thermodynamic information do we need to know to compute
the equilibrium constant of the reaction (a) at 298 K and (b) at
350 K? (You may assume that no phase transitions take place in this
temperature range.)

Part A
Calculate the standard enthalpy change for the reaction
2A+B⇌2C+2D
where the heats of formation are given in the following
table:
Substance
ΔH∘f
(kJ/mol)
A
-227
B
-399
C
213
D
-503
Express your answer in kilojoules.
Answer= 273kJ
Part B:
For the reaction given in Part A, how much heat is absorbed when
3.70 mol of A reacts?
Express your answer numerically in kilojoules.
Part C:
For the reaction given in Part A, ΔS∘rxn is 25.0 J/K ....

1.
Calculate the standard free energy change at 500 K for the
following reaction.
Cu(s) +
H2O(g) à CuO(s) +
H2(g)
ΔH˚f
(kJ/mol)
S˚
(J/mol·K)
Cu(s)
0
33.3
H2O(g)
-241.8
188.7
CuO(s)
-155.2
43.5
H2(g)
0
130.6
2. When
solid ammonium nitrate dissolves in water, the resulting solution
becomes cold. Which is true and why?
a. ΔH˚
is positive and ΔS˚ is positive
b. ΔH˚
is positive and ΔS˚...

PART 1. Which of the following reactions are spontaneous
(favorable). Check all that apply.
A. 2Mg(s)+O2(g)--->2MgO(s) delta G=-1137kj/mol
B.NH3(g)+HCl(g)--->NH4Cl(s) delta G=-91.1 kj/mol
C.AgCl(s)--->Ag+(aq)+Cl-(aq) delta G=55.6 kj/mol
D.2H2(g)+O2(g)--->2H2O(g) delta G=456 kj/mol
E.C(s)+H2O(l)--->CO(g)+H2(g) delta G=90.8 kj/mol
F.CH4(g)+2O2(g)--->CO2(g)+2H2O(l) delta G=-820
kj/mol
PART 2. Calculate the
standard entropy, ΔS°rxn, of the following reaction at 25.0 °C
using the data in this table. The standard enthalpy of the
reaction, ΔH°rxn, is –633.1 kJ·mol–1.
3C2H2(g)--->C6H6(l) ΔS°rxn=____JxK-1xmol-1
Then calculate Gibbs free energy for ΔG°rxn in kjxmol-1
Finally,...

Thermodynamics: Consider the equilibrium reaction A(g) + B(g)
-><- C(g)+D(g). At T=298 K, the standard enthalpies of
formation of the components in the gas phase are -20, -40, -30, and
-10 kJ/mol for A,B,C, and D, respectively. The standard-state
entropies of the components in the gas phase are 30, 50, 50, and 80
J/(mol K), in the same order. The vapor pressure of liquid C at
this temperature is 0.1 bar, while all other components are
volatile gases with Henry's...

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