Question

A) A student determines the value of the equilibrium constant to
be **2.57×10 ^{14}** for the following
reaction.

Based on this value of K

delta G° for this reaction is expected to be (greater, less) ________ than zero.

Calculate the free energy change for the reaction of

delta G°

B) A student determines the value of the equilibrium constant to
be **3.81×10 ^{-18}** for the following
reaction.

Based on this value of K

delta G° for this reaction is expected to be (greater, less) ______ than zero.

Calculate the free energy change for the reaction of

deltaG°

C) A student determines the value of the equilibrium constant to
be **1.87×10 ^{-18}** for the following
reaction.

**N _{2}(g)** +

Based on this value of K

delta G° for this reaction is expected to be (greater,less) _____ than zero.

Calculate the free energy change for the reaction of

delta G°

Answer #1

1)

2HBr(g) + Cl2(g) --> 2HCl(g) + Br2(g) K = 2.57×1014

since K > 1, this must favours products

dG < 1

dG = -RT*ln(K)

dG = -8.314*298*ln(2.57*10^14) = -82206.079 J = -82.20 kJ

this is based in 2 mol of HBr so..

for

**1.81** moles of **HBr(g)**

-82.20 * 1.81/2 = **-74.391 kJ**

2)

Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g) K = 3.81×10-18

since K < 1 , this must favours reactants

dG > 0

dG = -RT*ln(K)

dG = -8.314*298*ln(3.81*10^-18) = 99372.6937 J/mol per 1 mol of Fe3O4

we need 2.08 so--> 2.08*99372.6937 = 206695.20 J = 206.69 kJ

3)

K = 1.87*10^-18

dG!° = -RT*ln(K)

dG° = -8.314*298*ln(1.87*10^-18) = 101135.95 J

this is per 1 mol of N2

so for 2.27:

2.27*101135.95 = 229578.6065 J = 229.578 kJ

A student determines the value of the equilibrium constant to be
1.43×1064 for the following reaction. 2Na(s) + 2H2O(l)2NaOH(aq) +
H2(g) Based on this value of Keq: G° for this reaction is expected
to be (greater, less) than zero. Calculate the free energy change
for the reaction of 2.31 moles of Na(s) at standard conditions at
298K. G°rxn = kJ

� Gibbs Free Energy: Equilibrium Constant
Nitric oxide, NO, also known as nitrogen monoxide, is one of the
primary contributors to air pollution, acid rain, and the depletion
of the ozone layer. The reaction of oxygen and nitrogen to form
nitric oxide in an automobile engine is
N2(g)+O2(g)?2NO(g)
The spontaneity of a reaction can be determined from the free
energy change for the reaction, ?G?.
A reaction is spontaneous when the free energy change is less
than zero.
A reaction...

Consider the reaction: CO(g) + Cl2(g)COCl2(g) Using standard
thermodynamic data at 298K, calculate the free energy change when
1.900 moles of CO(g) react at standard conditions. G°rxn = ?????
kJ

Debate continues on the practicality of H2 gas as a fuel. The
equilibrium constant for the reaction CO(g)+ H2O (g) <-->
CO2(g)+ H2 (g) is 1.0x10^5 at 25 degrees C. Calculate the standard
free energy change (R=8.134 j/k*mol) and, without doing any
calculations, estimate delta H(rxn) and delta S(rxn)

Item 5
The equilibrium constant of a system, K, can be related
to the standard free energy change, ΔG, using the
following equation:
ΔG∘=−RTlnK
where T is standard temperature in kelvins and
R is the gas constant.
Under conditions other than standard state, the following
equation applies:
ΔG=ΔG∘+RTlnQ
In this equation, Q is the reaction quotient and is
defined the same manner as K except that the
concentrations or pressures used are not necessarily the
equilibrium values.
Part A
Acetylene,...

Use the free energies of formation given below to calculate the
equilibrium constant (K) for the following reaction at 298 K.
2 HNO3(aq) + NO(g) → 3 NO2(g) +
H2O(l) K = ?
ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1
Calculate the ΔG∘rxn for the reaction using the following
information.
4HNO3(g)+5N2H4(l)→7N2(g)+12H2O(l)
ΔG∘f(HNO3(g)) = -73.5 kJ/mol;
ΔG∘f(N2H4(l)) = 149.3 kJ/mol;
ΔG∘f(N2(g)) = 0 kJ/mol;
ΔG∘f(H2O(l)) = -273.1 kJ/mol.
Calculate the ΔG°rxn using the following
information.
2 H2S(g) + 3 O2(g) → 2...

a) Calculate the equilibrium constant at 17 K for a reaction
with ΔHrxno = 10 kJ and ΔSrxno = -100 J/K. (Don't round until the
end. Using the exponent enlarges any round-off error.)
b) Calculate the equilibrium constant at 146 K for the
thermodynamic data in the previous question (Notice that Keq is
larger at the larger temperature for an endothermic reaction)
c) Calculate the equilibrium constant at 43 K for a reaction
with ΔHrxno = 10 kJ and
ΔSrxno...

Consider the following system at equilibrium where H° = -111 kJ,
and Kc = 0.159, at 723 K.
N2(g) + 3H2(g) --> 2NH3(g) When 0.32 moles of H2(g) are
removed from the equilibrium system at constant temperature:
1- the value of Kc
A. increases.
B. decreases.
C. remains the same.
2- the value of Qc
A. is greater than Kc.
B. is equal to Kc.
C. is less than Kc. t
3- the reaction must:
A. run in the forward...

For a gaseous reaction, standard conditions are 298 K and a
partial pressure of 1 atm for all species. For the reaction
N2(g)+3H2(g)==<>2NH3(g) the standard change in Gibbs free
energy is Delta G=-72.6 kJ/mol. What is Delta G for this reaction
at 298 K when the partial pressures are
P N2 = 0.200 atm, P H2 = 0.150 atm, and P NH3 = 0.900 atm

The change in Gibbs free energy determines the direction in
which a reaction proceeds. what kind of value does gibbs free
energy have for the forward reaction to be spontaneous? gibbs free
energy is also related to the reactions equilibrium constant. from
what values can we calculate the equilibrium constant?

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