End of Module 13 Quiz

2 NH 3 (g)  <—–> N 2 (g) + 3 H 2 (g)

2 KCl (s) + 3 O 2 (g)  <—–> 2 KClO 3 (s)

2 H 2 O (l)  <—–>    H 3 O + (aq) +    OH – (aq)

2 CO 2 (g)    <—–> 2 CO (g)     +    O 2 (g)

2 Li  + (aq) + CO 3 2- (aq)    <—–>  Li 2 CO 3 (s)

Given the equilibrium concentrations of the reactants and products, calculate the K C (the
chemical equilibrium constants for the chemical equilibrium reactions:
2 NH 3 (g)  <—–>  N 2 (g) + 3 H 2 (g)
Equilibrium concentrations: [ N2 ] = 0.0200 M, [ H2 ] = 0.0200 M and [ NH3 ] = 0.0100 M

2 KCl (s) + 3 O 2 (g)  <—–>  2 KClO 3 (s)

Equilibrium concentrations: [ O 2 ] = 0.025 M

2 H2O (l)  <—–>    H 3 O + (aq) +     +   OH – (aq)
Equilibrium concentrations: [ H 3 O + ] = 1.0 X 10 – 8 M,  [ OH – ] = 1.0 X 10 – 6 M

2 CO 2 (g)    <—–>  2 CO (g)     +    O 2 (g)

Equilibrium concentrations: [ CO2 ] = 3.0 M, [ CO ] = 2.0 M, [ O2 ] = 1.5 M

2 Li + (aq) + CO 3 2- (aq)    <—–>  Li 2 CO 3 (s)
Equilibrium concentrations: [ Li + ] = 0.20 M, [ CO 3 2- ] = 0.10 M

Calculate the K C for the coupled reactions:

6. 2 CO2 (g)  + H2O (g)    <—–>  2 O2 (g)  + CH2CO (g)                     K C 1 = 6.1 x 10 8
7. CH4 (g) + 2 O2 (g) <—–> CO2 (g)  + 2 H2O (g)                               K C 2 = 1.2 x 10 14

——————————————————————————-

CH4 (g) + 2 CO2 (g)    <—–>     CH2CO (g) + H2O (g)

What is numerical value of K C for the reaction:

3 N2 (g) + 3 O2 (g) <—–> 6 NO (g)         K C = ?

2 NO (g)    <—–> N2 (g)  + O2 (g)             K’ C =  3.5 x 10 – 6

Calculate the value of K C for the coupled reaction below:

CaCO3 (s)    <—–>    Ca2+ (aq)     +      CO3 2- (aq)           K C 1 = 10 – 6.3

HCO3 – (aq)    <—–>      H+ (aq)     +      CO3 2- (aq)           K C 2 = 10 – 10.3

Calculate the K C for the coupled chemical equilibria below: 

2  NO (g)   <—–>   N2(g)  +  O2 (g)                        K C1 = 1 x 10 30

N2 (g)  + Br2 (g)  +  O2 (g)  <—–> 2 NOBr (g)     K C2 = 2 x 10 – 27

If the reaction below is initially at equilibrium, and then each of the following changes are made, predict which direction the reaction rate will be fastest until equilibrium is once again established: forward, reverse, or no change.

2 CO (g)  + O2 (g)    <—–>  2 CO2 (g)

The amount of O2 is increased:

If the reaction below is initially at equilibrium, and then each of the following changes are made, predict which direction the reaction rate will be fastest until equilibrium is once again established: forward, reverse, or no change.

C (s) + H2O (g) <—–> CO (g) + H2 (g)

Removing carbon:

If the reaction below is initially at equilibrium, and then each of the following changes are made, predict which direction the reaction rate will be fastest until equilibrium is once again established: forward, reverse, or no change.

PCl5 (g)  <—–>  PCl3 (g) + Cl2 (g)

adding Cl2:

What would happen to the position of the equilibrium when the following changes are made to the equilibrium system below?

CH4 (g)  + 2 H2S (g) <—–>     CS2 (g) + 4 H2 (g)

Decrease the concentration of dihydrogen sulfide (hydrosulfuric acid).

What would happen to the position of the equilibrium when the following changes are made to the equilibrium system below?

CH4 (g)  + 2 H2S (g) <—–>     CS2 (g) + 4 H2 (g)

Increase the pressure on the system.

What would happen to the position of the equilibrium when the following changes are made to the equilibrium system below?

CH4 (g)  + 2 H2S (g) <—–>     CS2 (g) + 4 H2 (g)

Increase the temperature of the system.

What would happen to the position of the equilibrium when the following changes are made to the equilibrium system below?

CH4 (g)  + 2 H2S (g) <—–>     CS2 (g) + 4 H2 (g)

Adding a catalyst

What would happen to the position of the equilibrium when the following changes are made to the equilibrium system below?

2 SO3 (g)  <—–>  2 SO2 (g)    +    O2 (g)

Increasing the volume of the system

What would happen to the position of the equilibrium when the following changes are made to the reaction below?

2 HgO (s) <—–>  Hg (s)    + O2 (g)

Increasing the pressure of the system

When the volume of the following mixture of gases is increased, what will be the effect on the equilibrium position?

4 HCl (g)    +  O2 (g) <—–>   2 H2O (g)  +  2 Cl2 (g)

Predict the effect of decreasing the volume of the container for each equilibrium.

2 H2O (g) +  N2 (g) <—–>  2H2(g) + 2NO(g)

Predict the effect of decreasing the volume of the container for each equilibrium

 SiO2 (g)  + 4 HF  <—–>     SiF4 (g) + 2 H2O(g)

Predict the effect of decreasing the volume of the container for each equilibrium

 CO (g) + H2 (g) <—–>  C (s) + H2O (g)

Predict the effect of decreasing the temperature on the position of the following equilibrium.

H2 (g) +  Cl2(g)  <—–> 2 HCl (g)  + 49.7 kJ

 Predict the effect of decreasing the temperature on the position of the following equilibrium

 SiO2 (g)  + 4 HF + Heat  <—–>     SiF4 (g) + 2 H2O(g)

Predict the effect of decreasing the temperature on the position of the following equilibrium

 CO (g)  + H2O (g)  <—–>     CO2 (g)    +  H2 (g)  + Heat

At a given temperature, the Keq for the chemical equilibrium reaction:

2 HI (g) <—–> H2 (g) + I2 (g)                K C = 1.4 x 10 – 2

If the concentrations of both H2 (g) and I2 (g) at equilibrium are 2.00 x 10 – 4. Find the equilibrium concentration of HI (g). 

 Acetic acid dissociates in water. If K C = 1.80 x 10 -5 and the equilibrium concentrations of acetic acid is 0.09986 M, what is the concentration of H+ (aq) and CH3COO –(aq)?

CH3COOH (aq)    <—–>    H+ (aq)  + CH3COO – (aq)

 At 60.2 oC the equilibrium constant for the reaction: N2O4 (g) <—–>  2 NO2 (g)

Is 8.75 x 10 -2. At equilibrium at this temperature a vessel contains N2O4 at concentration of 1.72 x 10 -2 M. What concentration of NO2 does it contain?

The table below show non-equilibrium concentrations of reactions and products for the following reaction at 527°C. The value of K C for this reaction at 527°C is 5.10

CO(g)  + H2O(g)   <—–>  H2 (g)   +   CO2 (g)

From the above data, calculate the reaction quotient Q and predict in which direction the reaction will proceed.

The table below shows non-equilibrium concentrations of reactions and products for the following reaction at 1000°C. The value of Kc for this reaction at 1000°C is 1.00 x 10-13

2 HF(g) <—–>  H2(g)  +  F2(g)

From the above data, calculate the reaction quotient Q and predict in which direction the reaction will proceed.

For the reaction below:  K C = 19.9 

Cl2 (g)  + F2 (g) <—–>  2 ClF(g)

Predict the shift of the chemical equilibrium, if the equilibrium concentrations of reactants and the product are given below:

[ Cl2 ] = 0.40 M, [ F ] = 0.20 M and [ ClF ] = 7.4 M

The value of K C at 25 oC for the equilibrium reaction: C (graphite) + CO2 (g) <—–> 2 CO (g) is 3.7 x 10 -23. Predict the shift of the chemical equilibrium if 2.0 mol of CO and 2.0 mol of CO2 are mixed in 1.0 L container with a suitable catalyst to make the reaction progress at this temperature.

The concentration equilibrium constant K C for the gas-phase reaction H2CO <—–>  H2 + CO has the

numerical value 0.50 at a given temperature. A mixture of H2CO, H2, and CO is introduced into a flask at this temperature. After a short time, analysis of a small sample of the reaction mixture shows the concentration to be [H2CO] = 0.50 M [H2] = 1.50 M and [CO] = 0.25 M Classify each of the following statements about this reaction as true or false

Nitrogen dioxide is introduced into a flask at a pressure of one atmosphere (1.00 atm). After some time, it dimerizes to produce N₂O₄. Use ICE.

What is the equilibrium partial pressure of N₂O₄ at a temperature where the equilibrium constant for dimerization is K p = 3.33?

2 NO₂ (g)  <—–>   N₂O₄ (g)

When 2.00 mol each of hydrogen and iodine are mixed in a 1.00­L flask, 3.50 mol of HI is produced at equilibrium

H₂ (g)  + I₂ (g) <—–>   2 HI (g)

The equilibrium concentration x = 1.75 M {3.50 mol /2]

What is the value of the equilibrium constant K C for this reaction.

Write an ICE table (but don’t solve it) for 0.66 atm H₂ reacting with 1.28 atm O₂ according to the equation:

2 H₂ (g) + O₂ (g) <—–> 2 H₂O (g)

In a container of 10.0 L volume, I mix 1.0 mol N₂, 1.0 mol H₂ and 0.5 mol O₂. Write an ICE table (but don’t solve it) for the equilibrium:

N₂ (g) + 4 H₂ (g) +  O₂ (g) <—–> N₂H₄ (g) + 2 H₂O (g)

In the reaction 2 NO₂ (g) <—–> N₂O₄ (g) the initial concentration of N₂O₄ was 0.100 M and NO₂ was 0.000 M. At equilibrium, the concentration of N₂O₄ was measured as 0.009 M. What is the value Kc. (First set up an ICE table)

In the reaction 2 NO₂ (g) <—–>  N₂O₄ (g) the initial concentration of NO₂ was 0.250 M and N₂O₄ was 0.000 M. At equilibrium, the concentration of N₂O₄ was measured as 0.0133 M

What is the value KC?

Initially, a mixture of 0.100 M NO, 0.050 M H₂ and 0.100 M H₂O was allowed to reach equilibrium. There was no N₂ present initially. At equilibrium, the concentration of NO was found to be 0.062 M. Calculate Kc,

2 NO (g) + 2 H₂ (g) <—–>  N₂ (g)    + 2 H₂O (g)

In the following reaction, K p = 9.3×10 ^-7 at room temperature. Calculate the equilibrium concentration of N₂O₄ in a flask initially containing only 3.00 atm of NO₂

2 NO₂ (g)  <—–>  N₂O₄ (g)

Iodine molecules dissociate at high temperature according to the reaction

I₂ (g) <—–> 2 I (g)

If KP = 4.5xl0 ^{– 4} and the reaction initially starts with only I₂ (g) with a pressure of 1.000 atm, what is the pressure of (a) I₂ (g) at equilibrium?

0.05 mol H₂(g) and 0.05 mol Br₂(g) are placed together in a 5.0 L flask and heated to 700 K. What is the concentration of each substance in the flask at equilibrium if Kc= 64 at 700 K?

H₂ (g)  +      Br₂ (g) <—–>  2 HBr (g)

In the following chemical equilibrium, 1.0 mol of NO (g), Cl₂ mol were added to a 1 L container. As a result of the reaction:

2 NO (g) + Cl₂ (g)  <—–>   2 NOCl (g)

The equilibrium concentration of NOCl became 0.96 M. What is the value of Kc?

Write an ICE table (but don’t solve it) for 1.0 M SO3 reacting with 2.0 M H2O according to the equation: 

SO3 (g) + H2O (g) <—–> H2SO4 (g)