Substitute the equilibrium concentrations into the equilibrium-constant. expression and solve for x: If you have an equation- solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic in x: Thus, solving the quadratic equation leads to two solutions for x: x = Kc = [HI]² (2x)² [H₂][1₂] (1.000 - x)(2.000 - x) X = = 4x² 46.5x2 – 151.5x + 101.0 0 = For a general quadratic equation of the form: ax² + bx + c = 0 The solution takes the form (two roots): -b ± √b²-4(a)(c) 50.5(x² - 3.000x + 2.000) 2a -(-151.5) + V1(-151.5)² - 4(46.5)(101.0) 2(46.5) Substitute x = 2.323 into the expressions for the equilibrium concentrations of H₂ and 12. Reject negative solutions or concentrations. THUS: use x = 0.935 to find the equilibrium concentrations: = 50.5 - 2.323 or 0.935 [H₂] = 1.000 x = 0.065 M [1₂] = 2.000- x = 1.065 M [HI] = 2x = 1.87 M A 1.000-L flask is filled with 1.000 mol of H₂ and 2.000 mol of 12 at 448 °C. The value of the equilibrium constant K, for the reaction: H₂(g) + 12(g) = 2 HI(g) C at 448 °C is 50.5. What are the equilibrium concentrations of H₂, 12₂, and HI in moles per liter? Solution Plan: In this case, we are not given any of the equilibrium concentrations. We must form relationships that relate the initial concentrations to those at equilibrium. The procedure is similar to that previously outlined, we just have less initial information. First, we note the initial concentrations of H₂ and 12 in the 1.000-L flask: Second, we construct a table in which we tabulate the initial concentrations: [H₂] = 1.000 M and [1₂] = 2.000 M H₂(g) Initial Change Equilibrium 1.000 M + 12(8) 2.000 M 2 HI(g) OM
Substitute the equilibrium concentrations into the equilibrium-constant. expression and solve for x: If you have an equation- solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic in x: Thus, solving the quadratic equation leads to two solutions for x: x = Kc = [HI]² (2x)² [H₂][1₂] (1.000 - x)(2.000 - x) X = = 4x² 46.5x2 – 151.5x + 101.0 0 = For a general quadratic equation of the form: ax² + bx + c = 0 The solution takes the form (two roots): -b ± √b²-4(a)(c) 50.5(x² - 3.000x + 2.000) 2a -(-151.5) + V1(-151.5)² - 4(46.5)(101.0) 2(46.5) Substitute x = 2.323 into the expressions for the equilibrium concentrations of H₂ and 12. Reject negative solutions or concentrations. THUS: use x = 0.935 to find the equilibrium concentrations: = 50.5 - 2.323 or 0.935 [H₂] = 1.000 x = 0.065 M [1₂] = 2.000- x = 1.065 M [HI] = 2x = 1.87 M A 1.000-L flask is filled with 1.000 mol of H₂ and 2.000 mol of 12 at 448 °C. The value of the equilibrium constant K, for the reaction: H₂(g) + 12(g) = 2 HI(g) C at 448 °C is 50.5. What are the equilibrium concentrations of H₂, 12₂, and HI in moles per liter? Solution Plan: In this case, we are not given any of the equilibrium concentrations. We must form relationships that relate the initial concentrations to those at equilibrium. The procedure is similar to that previously outlined, we just have less initial information. First, we note the initial concentrations of H₂ and 12 in the 1.000-L flask: Second, we construct a table in which we tabulate the initial concentrations: [H₂] = 1.000 M and [1₂] = 2.000 M H₂(g) Initial Change Equilibrium 1.000 M + 12(8) 2.000 M 2 HI(g) OM
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 51QRT: At room temperature, the equilibrium constant Kc for the reaction
2 NO(g) ⇌ N2(g) + O2(g)
is 1.4 ×...
Related questions
Question
Can you explain the concept? How do I do the calculations? For example, where did the 4x2 came from?
![Substitute the equilibrium
concentrations into the
equilibrium-constant.
expression and solve for x:
If you have an equation-
solving calculator, you can
solve this equation directly
for x. If not, expand this
expression to obtain a
quadratic in x:
Thus, solving the quadratic
equation leads to two
solutions for x:
x =
Kc
=
[HI]²
(2x)²
[H₂][1₂] (1.000 - x)(2.000 - x)
X =
=
4x²
46.5x2 – 151.5x + 101.0 0
=
For a general quadratic equation of the form:
ax² + bx + c = 0
The solution takes the form (two roots):
-b ± √b²-4(a)(c)
50.5(x² - 3.000x + 2.000)
2a
-(-151.5) + V1(-151.5)² - 4(46.5)(101.0)
2(46.5)
Substitute x = 2.323 into the expressions for
the equilibrium concentrations of H₂ and 12.
Reject negative solutions or concentrations.
THUS: use x = 0.935 to find the equilibrium
concentrations:
=
50.5
-
2.323 or 0.935
[H₂] = 1.000 x = 0.065 M
[1₂] = 2.000- x = 1.065 M
[HI] = 2x = 1.87 M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637a8243-98d2-45ca-8dd1-4e9d7d0f97ca%2Ff0379667-69a2-4114-bf16-323322b45833%2Fwqletfo.jpeg&w=3840&q=75)
Transcribed Image Text:Substitute the equilibrium
concentrations into the
equilibrium-constant.
expression and solve for x:
If you have an equation-
solving calculator, you can
solve this equation directly
for x. If not, expand this
expression to obtain a
quadratic in x:
Thus, solving the quadratic
equation leads to two
solutions for x:
x =
Kc
=
[HI]²
(2x)²
[H₂][1₂] (1.000 - x)(2.000 - x)
X =
=
4x²
46.5x2 – 151.5x + 101.0 0
=
For a general quadratic equation of the form:
ax² + bx + c = 0
The solution takes the form (two roots):
-b ± √b²-4(a)(c)
50.5(x² - 3.000x + 2.000)
2a
-(-151.5) + V1(-151.5)² - 4(46.5)(101.0)
2(46.5)
Substitute x = 2.323 into the expressions for
the equilibrium concentrations of H₂ and 12.
Reject negative solutions or concentrations.
THUS: use x = 0.935 to find the equilibrium
concentrations:
=
50.5
-
2.323 or 0.935
[H₂] = 1.000 x = 0.065 M
[1₂] = 2.000- x = 1.065 M
[HI] = 2x = 1.87 M
![A 1.000-L flask is filled with 1.000 mol of H₂ and 2.000 mol of 12 at 448 °C. The value
of the equilibrium constant K, for the reaction: H₂(g) + 12(g) = 2 HI(g)
C
at 448 °C is 50.5. What are the equilibrium concentrations of H₂, 12₂, and HI in moles
per liter?
Solution
Plan: In this case, we are not given any of the equilibrium concentrations. We
must form relationships that relate the initial concentrations to those at
equilibrium. The procedure is similar to that previously outlined, we just have
less initial information.
First, we note the initial
concentrations of H₂ and 12
in the 1.000-L flask:
Second, we construct a
table in which
we tabulate the initial
concentrations:
[H₂] = 1.000 M and [1₂] = 2.000 M
H₂(g)
Initial
Change
Equilibrium
1.000 M
+
12(8)
2.000 M
2 HI(g)
OM](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637a8243-98d2-45ca-8dd1-4e9d7d0f97ca%2Ff0379667-69a2-4114-bf16-323322b45833%2Fgpvrrbo.jpeg&w=3840&q=75)
Transcribed Image Text:A 1.000-L flask is filled with 1.000 mol of H₂ and 2.000 mol of 12 at 448 °C. The value
of the equilibrium constant K, for the reaction: H₂(g) + 12(g) = 2 HI(g)
C
at 448 °C is 50.5. What are the equilibrium concentrations of H₂, 12₂, and HI in moles
per liter?
Solution
Plan: In this case, we are not given any of the equilibrium concentrations. We
must form relationships that relate the initial concentrations to those at
equilibrium. The procedure is similar to that previously outlined, we just have
less initial information.
First, we note the initial
concentrations of H₂ and 12
in the 1.000-L flask:
Second, we construct a
table in which
we tabulate the initial
concentrations:
[H₂] = 1.000 M and [1₂] = 2.000 M
H₂(g)
Initial
Change
Equilibrium
1.000 M
+
12(8)
2.000 M
2 HI(g)
OM
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