Let's start with one mole of pure NH3, given that the pressure is 5 bars
Let's start with one mole of pure NH3, given that the initial pressure is 1 bar and the final
equalities pressure is 5 bars
2NH3(g) ----> N2 (g) + 3H2 (g)
Final Partial pressure Kp
𝑃𝐻2 ∗ (𝑃𝑁2 )4
(3 ∗ 2) ∗ (24 )
(1 − 2(2))
6 ∗ 16
Let's start with one mole of pure CH3 OH, given that the initial pressure is 1 bar and the final
equalities pressure is 50 bars
CO(g) + 2H2 (g) ----> CH3 OH(g)
Po -x + Po -2x + x = 50
Po -x = 50/2 = 25
Po -2x = 25 – x
Final Partial pressure Kp
(𝑃𝐶𝐻3 𝑂𝐻 )2
(𝑃𝐶𝑂 )2 ∗ 𝑃𝐻2
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Running head: ENTROPY CHANGE
Name of Author
When calculating the Standard Gibbs free energy of formation of a certain compound,
there various steps needed to be undertaken. These are standard enthalpy of formation which
is designated as (ΔHƒ°), and absolute standard entropy which is designated as (ΔS°) and
standard temperature which is (298 K).
ΔGƒ° = ΔHƒ° (given compound) - TΔS° (given compound)
In formation of mercury liquid from vapour, the product is liquid while the vapour as
reactants in a reverse equation. Therefore, the given values above of standard Gibbs free
energy of formation will be used in calculating the formation of free energy and enthalpy
ΔG° (reaction) = ΣGƒ° (products) − ΣGƒ° (reactants)
Therefore energy of formation is 61.4 Kj/mol + 31.8Kj/mol=93.2kJ/mol
While the enthalpy of formation is 29.6kJ/mol.
Change in Hvap=H vapor-H liquid
Therefore since we have Hvapor as 61.4 Kj/mol and that of liquid is 93.2kJ/mol. computing the
two values that is;
Is 61.4 Kj/mol-93.2kJ/mol=-31.8 Kj/mol.
Therefore for Gibbs energy of vaporization is
31.8 Kj/mol-93.2kJ/mol= -61.4 Kj/mol
Vaporization of mercury equation is given below. Mercury is liquid in nature.
The equilibrium constant is; 31.8kJ/mol. this is equal to formation of mercury vapour at
constant pressure and temperature
The equilibrium constant of an equation is given by K which symbolizes the value of Q
when the overall reaction of the equation is at equilibrium. K uniquely has a value at a given
reaction equation at constant pressure and temperature.
both Q and K are expressed in terms of concentrations, and partial pressure.
Calculating the vapor pressure equilibrium is given by In P1/P2=change H vap /R
Assuming it is an ideal gas, PV=NRT and an activity of 1.
Moles of liquid mercury = 2
P= (2×8.3142×298) ÷1=4955.2632
Therefore, P1= 100000pascals and p2 is 4955pascals
In (100000÷4955.26) ÷ (1/273...
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