# 13.3 Shifting equilibria: le châtelier’s principle  (Page 6/10)

 Page 6 / 10

## Chemistry end of chapter exercises

The following equation represents a reversible decomposition:
${\text{CaCO}}_{3}\left(s\right)⇌\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)$

Under what conditions will decomposition in a closed container proceed to completion so that no CaCO 3 remains?

The amount of CaCO 3 must be so small that ${P}_{{\text{CO}}_{2}}$ is less than K P when the CaCO 3 has completely decomposed. In other words, the starting amount of CaCO 3 cannot completely generate the full ${P}_{{\text{CO}}_{2}}$ required for equilibrium.

Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.

What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?

The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants' side.

What would happen to the color of the solution in part (b) of [link] if a small amount of NaOH were added and Fe(OH) 3 precipitated? Explain your answer.

The following reaction occurs when a burner on a gas stove is lit:
${\text{CH}}_{4}\left(g\right)+2{\text{O}}_{2}\left(g\right)⇌{\text{CO}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(g\right)$

Is an equilibrium among CH 4 , O 2 , CO 2 , and H 2 O established under these conditions? Explain your answer.

No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO 3 , from sulfur dioxide, SO 2 , and oxygen, O 2 , shown here. At high temperatures, the rate of formation of SO 3 is higher, but the equilibrium amount (concentration or partial pressure) of SO 3 is lower than it would be at lower temperatures.
$2{\text{SO}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{SO}}_{3}\left(g\right)$

(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?

(b) Is the reaction endothermic or exothermic?

Suggest four ways in which the concentration of hydrazine, N 2 H 4 , could be increased in an equilibrium described by the following equation:
${\text{N}}_{2}\left(g\right)+2{\text{H}}_{2}\left(g\right)⇌{\text{N}}_{2}{\text{H}}_{4}\left(g\right)\phantom{\rule{5em}{0ex}}\text{Δ}H=95\phantom{\rule{0.2em}{0ex}}\text{kJ}$

Add N 2 ; add H 2 ; decrease the container volume; heat the mixture.

Suggest four ways in which the concentration of PH 3 could be increased in an equilibrium described by the following equation:
${\text{P}}_{4}\left(g\right)+6{\text{H}}_{2}\left(g\right)⇌4{\text{PH}}_{3}\left(g\right)\phantom{\rule{5em}{0ex}}\text{Δ}H=110.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

(a) $2{\text{NH}}_{3}\left(g\right)⇌{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}\text{Δ}H=92\phantom{\rule{0.2em}{0ex}}\text{kJ}$

(b) ${\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2\text{NO}\left(g\right)\phantom{\rule{5em}{0ex}}\text{Δ}H=181\phantom{\rule{0.2em}{0ex}}\text{kJ}$

(c) $2{\text{O}}_{3}\left(g\right)⇌3{\text{O}}_{2}\left(g\right)\phantom{\rule{5em}{0ex}}\text{Δ}H=-285\phantom{\rule{0.2em}{0ex}}\text{kJ}$

(d) $\text{CaO}\left(s\right)+{\text{CO}}_{2}\left(g\right)⇌{\text{CaCO}}_{3}\left(s\right)\phantom{\rule{5em}{0ex}}\text{Δ}H=-176\phantom{\rule{0.2em}{0ex}}\text{kJ}$

(a) Δ T increase = shift right, Δ P increase = shift left; (b) Δ T increase = shift right, Δ P increase = no effect; (c) Δ T increase = shift left, Δ P increase = shift left; (d) Δ T increase = shift left, Δ P increase = shift right.

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3 isotopes
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how many isotopes does hydrogen have
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