15.2.4 Solving Problems Far from Equilibrium

Question 1

Solving Problems Far from Equilibrium

Answer 1

• To solve equilibrium problems for reactions beginning far from equilibrium, first determine what the concentrations or partial pressures would be near equilibrium.
• The calculated concentrations, once one reactant or product is completely consumed, are used in a modified ICE diagram.

Question 2

note

Answer 2

• To solve equilibrium problems for reactions beginning far from equilibrium, first determine what the concentrations or partial pressures would be near equilibrium.
• Problem: 0.90 atm water (H 2 O), 3.0 atm carbon monoxide (CO), and 4.2 atm hydrogen gas (H 2 ) are allowed to react at 600 K. If K p for this reaction is 1.8 x 10 –7 at 600 K, what will the partial pressure of H 2 be at equilibrium?
• The reaction quotient (Q) is much greater than the equilibrium constant (K p ). Therefore, the reaction will proceed from right to left to reach equilibrium.
• To find the partial pressure of H 2 at equilibrium, find the limiting reagent. Since 1.6 atm CO remain after all of the H 2 is consumed, H 2 is the limiting reagent.
• When all of the H 2 is consumed, 1.4 atm (0.0 atm + 1.4 atm) CH 4 , 2.3 atm (0.90 atm + 1.4 atm) H 2 O, and
  1.6 atm (3.0 atm – 1.4 atm) CO are present. These values are added to a modified ICE diagram in a line labeled
  “Stoichiometry”.
• The problem can then be solved in the same manner as problems beginning near equilibrium. The reaction is now very near equilibrium, so x will be small.
• Plugging the values into the expression for the equilibrium constant and assuming x is small yields a partial pressure of H 2 of 7.1 x 10 –3 atm.

Question 3

Consider the following ICE diagram problem. The initial concentrations of each component are indicated by the I values. Which statement about this reaction is not correct?
CH4(g) + H2O(g) ->

Answer 3

The fact that x is very small is a very helpful assumption that we can use to simplify the equation for this reaction.

Question 4

Consider the following ICE diagram problem. The initial concentrations of each component are indicated by the I values. What are the four S values for this reaction and the final partial pressure for H2, the limiting reagent for this reaction?
CH4(g) + H2O(g) ->

Answer 4

1.1 atm; 1.95 atm; 0.6 atm; 0 atm; PH2 = 8.6*10^-3 atm

Question 5

Consider the following reaction: CH4(g) + H2O(g) ->

Answer 5

The value for the equilibrium constant, Kp, at this point is 0 because there is no H2 gas. (As a result, the partial pressure contribution by the H2 gas is 0.)

Question 6

Consider the following ICE diagram problem. The initial concentrations of each component are indicated by the I values. What are the four S values for this reaction and the final partial pressure for CO, the limiting reagent for this reaction?
CH4(g) + H2O(g) ->

Answer 6

1.1 atm; 1.9 atm; 0 atm; 0.3 atm; PCO = 1.39*10^-5 atm

Question 7

Look at the following ICE diagram problem. The initial concentrations of each component are indicated by the I values. Which statement about this reaction is not correct?
CH4(g) + H2O(g) ->

Answer 7

At equilibrium, the hypothetical value for x must be close to 3.9.

Question 8

Consider the following reaction: CH4(g) + H2O(g) ->

Answer 8

Because there is no methane and the Kp value is so small, all of the products will be (nearly) consumed as the equation shifts to the left.

Question 9

Look at the following ICE diagram problem. Which gas is the limiting reagent? CH4(g) + H2O(g) ->

Answer 9

CO

Question 10

What is the limiting reagent in the reaction, 2N2O5(g) ->

Answer 10

NO2

Question 11

Look at the following partial ICE diagram for the reaction. CH4(g) + H2O(g) ->

Answer 11

CO

Question 12

Look at the following ICE diagram problem. Which statement about this reaction is not correct?
CH4(g) + H2O(g) ->

Answer 12

If you hypothetically shift the equilibrium all the way to the left, the concentration of CO and H2 will be close to 0.