GSS 13 Chemical Equilibrium/Le Chatelier's Principle

At constant temperature, a change in volume will NOT affect the moles of substance present in which of the following?

• H2 (g) + I2 (g) $\Longleftrightarrow$ 2 HI (g)

• CO (g) + Cl2 (g) $\Longleftrightarrow$ COCl2 (g)

• PCl5 (g) $\Longleftrightarrow$ PCl3 (g) + Cl2 (g)

• N2 (g) + 3 H2 (g) $\Longleftrightarrow$ 2 NH3 (g)

• Fe3O4 (s) + 4 H2 (g) $\Longleftrightarrow$ 3 Fe (s) + 4 H2O (g)

C (s) + H2O (g)  $\Longleftrightarrow$  CO (g) + H2O (g) endothermic

An equilibrium mixture of the reactants is placed in a sealed container at  $150\ \degree C$ . The amount of the products may be increased by which of the following changes?

• increasing the volume of the container

• raising the temperature of the container and increasing the volume of the container

• raising the temperature of the container

• increasing the volume of the container and adding 1 mol of C (s) to the container

Consider the following system at $100\degree C$ : 

 $N_2O_4\ \left(g\right)\ \Longleftrightarrow\ 2\ NO_2\ \left(g\right)\ \ \ K_p\ =\ 11$  

Predict the direction in which the reaction will occur to reach equilibrium, starting with 0.10 mol of N2O4 and 0.20 mol of NO2 in a 2.0-L container

• forward

• it's already at equilibrium

• reverse

For the system: $CO_2\ \left(g\right)\ +\ H_2\ \left(g\right)\ \Longleftrightarrow\ CO\ \left(g\right)\ +\ H_2O\ \left(g\right)$  

K is 0.64 at 900 K. Calculate the equilibrium partial pressures of all species, starting with

 $P_{CO_2}=P_{H_2}=1.00\ atm;\ P_{CO}=P_{H_2O}=0$  

•  $P_{CO_2}=P_{H_2}=0.18;\ P_{CO}=P_{H_2O}=0.72\ atm$  

•  $P_{CO_2}=P_{H_2}=0.44\ atm;\ P_{CO}=P_{H_2O}=0.56\ atm$  

•  $P_{CO_2}=P_{H_2}=0.56\ atm;\ P_{CO}=P_{H_2O}=0.44\ atm$  

•  $P_{CO_2}=P_{H_2}=0.72\ atm;\ P_{CO}=P_{H_2O}=0.18\ atm$  

For the system: $CO_2\ \left(g\right)\ +\ H_2\ \left(g\right)\ \Longleftrightarrow\ CO\ \left(g\right)\ +\ H_2O\ \left(g\right)$  

K is 0.64 at 900 K. Calculate the equilibrium partial pressures of all species, starting with

 $P_{CO_2}=2.00\ atm;\ P_{H_2}=1.00\ atm;\ P_{CO}=P_{H_2O}=0$  

•  $P_{CO_{ }}=P_{H_2O}=0.70\ atm;\ P_{CO_2}=1.50\ atm;\ P_{H_2}=0.10\ atm$  

•  $P_{CO_{ }}=P_{H_2O}=0.30\ atm;\ P_{CO_2}=0.70\ atm;\ P_{H_2}=0.20\ atm$  

•  $P_{CO_{ }}=P_{H_2O}=0.60\ atm;\ P_{CO_2}=1.40\ atm;\ P_{H_2}=0.40\ atm$  

•  $P_{CO_{ }}=P_{H_2O}=0.80\ atm;\ P_{CO2}=1.60\ atm;\ P_{H_2}=0.50\ atm$  

The system $3\ Z\ \left(g\right)\ +\ Q\ \left(g\right)\ \Longleftrightarrow\ 2\ R\ \left(g\right)$  

is at equilibrium when the partial pressure of Q is 0.44 atm. Sufficient R is added to increase the partial pressure of Q temporarily to 1.5 atm. When equilibrium is re-established, the partial pressure of Q could be which of the following? Select all that apply.

• 1.5 atm

• 1.2 atm

• 0.80 atm

• 0.44 atm

• 0.40 atm

For this chemical reaction: $\ A\ \left(g\right)\ +\ 2\ B\ \left(s\right)\ +\ 3\ C\ \left(l\right)\ \Longleftrightarrow\ 4\ D\ \left(g\right)\ +\ 2\ E\ \left(g\right)$ if we want to increase the ratio of desired products D and E at equilibrium compared to reactants, how might we do this? Assume the reaction progresses and it feels cold to touch...

• Increase pressure in the constant volume container

• Raise the temperature with a hot water bath

• Increase the volume by expansion in a piston

• Removing product E as soon as it forms

• Adding a catalyst 

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