have 2 questions regarding this example in the first underlined fraction why is NO not squared ? And the second is can you explain the second underlined green line how can I get this equation can you show me the workout for it Thank you
have 2 questions regarding this example in the first underlined fraction why is NO not squared ? And the second is can you explain the second underlined green line how can I get this equation can you show me the workout for it Thank you
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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I have 2 questions regarding this example in the first underlined fraction why is NO not squared ?
And the second is can you explain the second underlined green line how can I get this equation can you show me the workout for it
Thank you
![Exercises performed during practice lesson
1. 100 g of Air are heated to 2300°C in a closed container and the following equilibrium is established:
N2(g) + O2(g) → 2NO(g)
Calculate the %vol of NO at the equilibrium knowing that Kp= 0.00270 and for air the % wt composition
is (75.4% N₂, 23.3% O₂ and 1.30% Ar)
K =Kp
1
RT
Initial nN₂
n0₂ =
An
N2(g) + O2(g) → 2NO(g)
no₂
-X
-X
Equilibrium nN₂-x nO₂-x
X =
The An= 2-1-1 is
equal to zero so
the Kc=Kp
-b = √b² - 4ac
2a
+2x
+2x
= 0.728 mol
-
XNO =
K =
Calculate the number of moles of each component on the reactor by considering the weight composition of the mixture
23.3
1.30
100 × 100 9
X
100 x
100 9
32 g/mol
40 g/mol
Applying the definition of Kc and the definition of molarity,
the equation of K can be written as:
[nNO1²
[2x]²
[nN₂ -x][n0₂ -x]
x² (4K) = KnN₂nO₂-xK(nN₂ + n0₂)
nN₂ =
Kc =
a = 4-K
With b = K (n0₂ + nN₂)
c = KnN₂n0₂
75.4
100 x 100
9
= 2.69 mol
28 g/mol
nNO
₂x10₂nN₂ x 10₂
Substitute the values oxygen moles and nitrogen moles in order to resolve the equation of x (variation of moles at the
equilibrium)
nN₂ initial moles - x = (2.69 -0.0351)mol = 2.65 mol
n0₂= initial moles - x = (0.728 -0.0351)mol = 0.693 mol
nNO= 2x = 2 x 0.0351 = 0.0702 mol
x =
nNO
nNO
ntot no₂+ nN₂ + nNO + nAr
0.0351
n Ar =
-0.0374 Not acceptable
0.0702 mol
3.42 mol
= K
= 0.021 = 2.1%
= 0.0325 mol
Exercises performed during practice lesson
4. A reactor of volume 0.5 liters at 300°C (constant) contains three gasses A, B and C in equilibrium as
described by the following equation:
A (g) + B (g) = C(g)
At equilibrium the composition is 2.5 moles of A, 2.5 moles of B and 8 moles of C. Calculate the new
equilibrium composition (in moles) if the volume of the reactor is reduced (keeping the temperature
constant) to 0.1 litres.
16
31
✪](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2Ffbd46d13-568e-40cd-bb32-78592e6c998a%2Fmpl4xmo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercises performed during practice lesson
1. 100 g of Air are heated to 2300°C in a closed container and the following equilibrium is established:
N2(g) + O2(g) → 2NO(g)
Calculate the %vol of NO at the equilibrium knowing that Kp= 0.00270 and for air the % wt composition
is (75.4% N₂, 23.3% O₂ and 1.30% Ar)
K =Kp
1
RT
Initial nN₂
n0₂ =
An
N2(g) + O2(g) → 2NO(g)
no₂
-X
-X
Equilibrium nN₂-x nO₂-x
X =
The An= 2-1-1 is
equal to zero so
the Kc=Kp
-b = √b² - 4ac
2a
+2x
+2x
= 0.728 mol
-
XNO =
K =
Calculate the number of moles of each component on the reactor by considering the weight composition of the mixture
23.3
1.30
100 × 100 9
X
100 x
100 9
32 g/mol
40 g/mol
Applying the definition of Kc and the definition of molarity,
the equation of K can be written as:
[nNO1²
[2x]²
[nN₂ -x][n0₂ -x]
x² (4K) = KnN₂nO₂-xK(nN₂ + n0₂)
nN₂ =
Kc =
a = 4-K
With b = K (n0₂ + nN₂)
c = KnN₂n0₂
75.4
100 x 100
9
= 2.69 mol
28 g/mol
nNO
₂x10₂nN₂ x 10₂
Substitute the values oxygen moles and nitrogen moles in order to resolve the equation of x (variation of moles at the
equilibrium)
nN₂ initial moles - x = (2.69 -0.0351)mol = 2.65 mol
n0₂= initial moles - x = (0.728 -0.0351)mol = 0.693 mol
nNO= 2x = 2 x 0.0351 = 0.0702 mol
x =
nNO
nNO
ntot no₂+ nN₂ + nNO + nAr
0.0351
n Ar =
-0.0374 Not acceptable
0.0702 mol
3.42 mol
= K
= 0.021 = 2.1%
= 0.0325 mol
Exercises performed during practice lesson
4. A reactor of volume 0.5 liters at 300°C (constant) contains three gasses A, B and C in equilibrium as
described by the following equation:
A (g) + B (g) = C(g)
At equilibrium the composition is 2.5 moles of A, 2.5 moles of B and 8 moles of C. Calculate the new
equilibrium composition (in moles) if the volume of the reactor is reduced (keeping the temperature
constant) to 0.1 litres.
16
31
✪
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