1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Jobisdone [24]
3 years ago
8

Consider the following system at equilibrium: 2A(aq)+2B(aq)⇌5C(aq) Classify each of the following actions by whether it causes a

leftward shift, a rightward shift, or no shift in the direction of the net reaction.a. increase (b)
b. increase(a)
c. increase(c)
d. decrease(a)
e. decrease(b)
f. decrease(c)
g. double(a) and reduce (b) to one half
h. double both (b) and (c)
Chemistry
1 answer:
Alexandra [31]3 years ago
4 0

Answer:

  • <em>Rightwardshift</em>: (a), (b), (f) and (h)
  • <em>Leftwardshift</em>: (c), (d), and (e)
  • <em>No shift</em>: (g)

Explanation:

1. Balanced chemical equation (given):

      2A(aq)+2B(aq)\rightleftharpoons 5C(aq)

2. Equilibrium constant

The equilibrium constant is the ratio of the product of the concentrations of the products, at equilibrium, each raised to its stoichiometric coefficient, to the product of the concentrations of the reactants, at the equilibrium, each raised to its stoichiometric coefficient.

          K_{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}

<u>a. Increase [B]</u>

  • Rightward shift

Since, by assumption, the temperature of the reaction is the same, the equilibrium constant   K_{c} is the same, meaning that an increase in the concentration of the species B must cause a rightward shift to increase the concentration of the species C, such that the ratio expressed by the equilibrium constant remains unchanged.

<u>b. Increase [A]</u>

  • Rightward shift.

This is exactly the same case for the increase of [B], since it is in the same side of the equilibrium chemical equation.

c. Increase [C]

  • Leftward shift.

C is on the right side of the equilibrium equation, thus, following Le Chatelier's principle, an increase of its concentration must shift the reaction to the left to restore the equilibrium. Of course, same conclusion is drawn by analyzing the expression for  K_{c} : by increasing the denominator the numerator has to increase to keep the same value of  K_{c}

d. Decrease [A]

  • Leftward shift.

This is the opposite change to the case {b), thus it will cause the opposite effect.

e. Decrease[B]

  • Leftward shift.

This is the opposite to case (a), thus it will cause the opposite change.

f. Decrease [C]

  • Rightward shift.

This is the opposite to case (c), thus it will cause the opposite change.

g. Double [A] and reduce [B] to one half

  • No shift

You need to perform some calculations and determine the reaction coefficient, Q_c to compare with the equilibrium constant K_{c}.

The expression for Q_c has the same form of the equation for  K_{c}. but the it uses the inital concentrations instead of the equilibrium concentrations.

            Q{c}=\frac{[C(aq)]^5}{[A(aq)]^2\cdot [B(aq)]^2}

Doubling [A] and reducing  [B] to one half would leave the product of [A]² by [B]² unchanged, thus Q_c  will be equal to K_{c}.

When  Q_c  = K_{c} the reaction is at equilibrium, so no shift will occur.

h. Double both [B] and [C]

  • Rightward shift.

Again, using the expression for Q_c, you will realize that the [C] is raised to the fifth power (5) while [B] is squared (power 2). That means that Q_c will be greater than  K_{c}..

When   Q_c  > K_{c} the equilibrium must be displaced to the left some of the reactants will need to become products, causing the reaction to shift to the right.

<u>Summarizing:</u>

  • Rightwardshift: (a), (b), (f) and (h)

  • Leftwardshift: (c), (d), and (e)

  • No shift: (g)

You might be interested in
What turns blue litmus paper red and have a ph greater than 7?
Naddik [55]
The answer is acidic 
6 0
3 years ago
6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH
Vadim26 [7]

Answer:

a) pH = 9.14

b) pH = 8.98

c) pH = 8.79

Explanation:

In this case we have an acid base titration. We have a weak base in this case the pyridine (C₅H₅N) and a strong acid which is the HCl.

Now, we want the know the pH of the resulting solution when we add the following volume of acid: 0, 10 and 20.

To know this, we first need to know the equivalence point of this titration. This can be known using the following expression:

M₁V₁ = M₂V₂  (1)

Using this expression, we can calculate the volume of acid required to reach the equivalence point. Doing that we have:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ = 0.125 * 25 / 0.1 = 31.25 mL

This means that the acid and base will reach the equivalence point at 31.25 mL of acid added. So, the volume of added acid of before, are all below this mark, so we can expect that the pH of this solution will be higher than 7, in other words, still basic.

To know the value of pH, we need to apply the following expression:

pH = 14 - pOH  (2)

the pOH can be calculated using this expression:

pOH = -log[OH⁻]  (3)

The [OH⁻] is a value that can be calculated when the pyridine is dissociated into it's ion. However, as this is a weak acid, the pyridine will not dissociate completely in solution, instead, only a part of it will be dissociated. Now, to know this, we need the Kb value of the pyridine.

The reported Kb value of the pyridine is 1.5x10⁻⁹ so, with this value we will do an ICE chart for each case, and then, calculate the value of the pH.

<u>a) 0 mL of acid added.</u>

In this case, the titration has not begun, so the concentration of the base will not be altered. Now, with the Kb value, let's write an ICE chart to calculate the [OH⁻], the pOH and then the pH:

       C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.125                                0             0

e)        -x                                   +x           +x

c)      0.125-x                              x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.125-x --> Kb is really small, so we can assume that x will be very small too, and 0.125-x can be neglected to only 0.125, and then:

1.5x10⁻⁹ = x² / 0.125

1.5x10⁻⁹ * 0.125 = x²

x = [OH⁻] = 1.37x10⁻⁵ M

Now, we can calculate the pOH:

pOH = -log(1.37x10⁻⁵) = 4.86

Finally the pH:

pH = 14 - 4.86

<h2>pH = 9.14</h2>

<u>b) 10 mL of acid added</u>

In this case the titration has begun so the acid starts to react with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.010) = 1x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 1x10⁻³ = 2.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 2.125x10⁻³ / (0.025 + 0.010) = 0.0607 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.0607                             0             0

e)        -x                                   +x           +x

c)      0.0607-x                           x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.0607-x --> 0.0607

1.5x10⁻⁹ = x² / 0.0607

1.5x10⁻⁹ * 0.0607 = x²

x = [OH⁻] = 9.54x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(9.54x10⁻⁶) = 5.02

Finally the pH:

pH = 14 - 5.02

<h2>pH = 8.98</h2>

<u>c) 20 mL of acid added:</u>

In this case the titration it's almost reaching the equivalence point and the acid is still reacting with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.020) = 2x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 2x10⁻³ = 1.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 1.125x10⁻³ / (0.025 + 0.020) = 0.025 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.025                                0             0

e)        -x                                   +x           +x

c)      0.025-x                             x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.025-x --> 0.025

1.5x10⁻⁹ = x² / 0.025

1.5x10⁻⁹ * 0.025 = x²

x = [OH⁻] = 6.12x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(6.12x10⁻⁶) = 5.21

Finally the pH:

pH = 14 - 5.21

<h2>pH = 8.79</h2>
5 0
3 years ago
Classify the following statements as observation, inference, prediction, measuring or classifying
Nadusha1986 [10]

Answer:

ggghy I am a great and the other hand the wildcats their own is the name of the

8 0
3 years ago
Is water a good solvent choice for recrystallization of NaCl?
nika2105 [10]

Answer:

Water is not a good solvent choice.

Explanation:

While water is good solvent because of its polotiry. Water is not good for the recrystallization process becuase being a good recystallization solvent means that a compound must dissolves easily when the solvent is warm, but that is less soluble at room temperature or when cooled in an ice bath. Water has the dissolves when warm part down. But for the cooled part which is the most important it can not do.

7 0
3 years ago
Night vision glasses detect
lana [24]

night vision glasses detect energy

5 0
3 years ago
Other questions:
  • what type of bond is involved when a metal from groups 1 or 2 bond with a nonmetal from groups 16 or 17
    11·1 answer
  • An aqueous solution of iron(II) sulfate (FeSO4) is prepared by dissolving 2.00 g in sufficient deionized water to form a 200.00
    12·1 answer
  • What changes as you descend through the water column
    15·1 answer
  • Do protons have attraction for neutrons yes or no
    5·1 answer
  • Please help solving number 48.
    10·1 answer
  • What was the commonality between Becquerel and the Curies in terms of their work? polonium radium uranium cesium
    11·2 answers
  • All animals need oxygen gas (O2) for their primary cellular-level functioning. Inside the cell, O2 is split apart into oxygen at
    7·1 answer
  • A weather balloon is filled with helium that occupies a volume of 5.00x104L at 0.995atm and 32.0C. After it is released, it rise
    14·1 answer
  • Is viscosity and surface tension the same
    15·1 answer
  • If you have 1.5 grams of silver nitrate, how many grams of silver could<br> you produce?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!