Question

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)

Delta S rxn =

Then calculate the standard gibbs free energy of the reaction Delta G rxn

Delta G rxn=

Answer #1

3C2H2(g)==>C6H6(l)

ΔH°rxn, = ΔH°f products -ΔH°f reactans

= 49.08-3*227.48 = -633.36KJ

ΔS°rxn, = ΔS°f products -ΔS°f reactans

= 173.4 -3*200.9 = -429.4J/Mole-K = -0.4294Kj/mole-K

ΔG^{0} = ΔH-TΔS

=- 633.1-298*-0.4294 = -505.14KJ

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,...

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...

Calculate the standard enthalpy of reaction for 2 C(graphite) +
3 H2(g) C2H6(g) Given the following standard enthalpy of combustion
data, ∆H˚comb (C(graphite) = –393.5 kJ·mol–1 H2(g) + ½ O2(g) H2O(l)
∆H˚rxn = –285.8 kJ·mol–1 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)
∆H˚rxn = –3119.6 kJ·mol–1 (a) –84.6 kJ·mol–1 (b) 2440.2 kJ·mol–1
(c) –3799.0 kJ·mol–1 (d) –224.5 kJ·mol–1(e) not enough information
provided

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?

Given the following data, calculate, ΔH rxn, ΔS rxn, and Δ rxn,
at 25° C for the equilibrium describe by the chemical equation.
What direction is the spontaneity of this system?
Mg (s) +HCl (aq) <---> H2 (g) +
MgCl2 (aq)
Mg(s)
HCl(aq)
H2 (g)
MgCl2(aq)
ΔH°f (Kj/mol)
0
-167.2
0
-801.3
S°(J/(mol K)
130.7
56.5
32.7
-24.0

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 ....

For each of the following reactions, calculate ΔH∘rxn,
ΔS∘rxn, ΔG∘rxn at 25 ∘C. State whether or not the
reaction is spontaneous. If the reaction is not spontaneous, would
a change in temperature make it spontaneous? If so, should the
temperature be raised or lowered from 25 ∘C?
2NH3(g)→N2H4(g)+H2(g)
2KClO3(s)→2KCl(s)+3O2(g)

Use Hess's Law to calculate the enthalpy of reaction, ΔH rxn,
for the reaction in bold below given the following chemical steps
and their respective enthalpy changes. Show ALL work!
2 C(s) + H2(g) → C2H2(g) ΔH°rxn = ?
1. C2H2(g) + 5/2 O2(g) → 2CO2 (g) + H2O (l) ΔH°rxn = -1299.6
kJ
2. C(s) + O2(g) → CO2 (g) ΔH°rxn = -393.5 kJ
3. H2(g) + ½ O2(g) → H2O (l) ΔH°rxn = -285.8 kJ

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...

Calculate the entropy of vaporization, ΔSvap, for A(l) at 25.0
°C given that the boiling point of A is 72.45 °C, and the molar
heat capacity of A(l) is 116.45 J/(mol·K). Assume that the molar
heat capacity of A(g) is 56.5% of that of A(l).
Calculate the standard Gibbs free energy of vaporization,
ΔG°vap, at 25.0 °C.
Determine the equilibrium constant, K, for the vaporization at
191.0 °C.
Determine the equilibrium constant, K, for the vaporization at
191.0 °C.

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