Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Short Answer

Expert verified
a. The equilibrium will shift to the left when the volume is increased. b. The equilibrium will shift to the right when the volume is increased. c. There is no shift in the equilibrium when the volume is increased. d. The equilibrium will shift to the right when the volume is increased. e. The equilibrium will shift to the right when the volume is increased.

Step by step solution

01

a. Analyze moles of gas

For the first reaction, count the moles of gas on each side: Left side: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) has 1 + 3 = 4 moles of gas. Right side: \(2 \mathrm{NH}_{3}(g)\) has 2 moles of gas.
02

a. Predict the shift

Since there are more moles of gas on the left side, the equilibrium will shift to the left when the volume is increased.
03

b. Analyze moles of gas

For the second reaction, count the moles of gas on each side: Left side: \(\mathrm{PCl}_{5}(g)\) has 1 mole of gas. Right side: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) has 1 + 1 = 2 moles of gas.
04

b. Predict the shift

Since there are more moles of gas on the right side, the equilibrium will shift to the right when the volume is increased.
05

c. Analyze moles of gas

For the third reaction, count the moles of gas on each side: Left side: \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g)\) has 1 + 1 = 2 moles of gas. Right side: \(2 \mathrm{HF}(g)\) has 2 moles of gas.
06

c. Predict the shift

Since there are the same number of moles of gas on both sides of the reaction, there is no shift in the equilibrium when the volume is increased.
07

d. Analyze moles of gas

For the fourth reaction, count the moles of gas on each side: Left side: \(\mathrm{COCl}_{2}(g)\) has 1 mole of gas. Right side: \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) has 1 + 1 = 2 moles of gas.
08

d. Predict the shift

Since there are more moles of gas on the right side, the equilibrium will shift to the right when the volume is increased.
09

e. Analyze moles of gas

For the fifth reaction, count the moles of gas on each side: Left side: \(\mathrm{CaCO}_{3}(s)\) has 0 moles of gas (as it is a solid). Right side: \(\mathrm{CO}_{2}(g)\) has 1 mole of gas.
10

e. Predict the shift

Since there are more moles of gas on the right side, the equilibrium will shift to the right when the volume is increased.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
One of the most important concepts in chemical equilibrium is Le Chatelier's principle. This principle is crucial for predicting how a system at equilibrium reacts to changes in concentration, temperature, and volume. According to Le Chatelier's principle, if an external change is applied to a system at equilibrium, the system adjusts itself to counteract the change and a new equilibrium is established.

For instance, when a student examines the effect of increasing the volume of the reaction container, it is Le Chatelier's principle that allows them to predict how the position of equilibrium will shift. If increasing volume, which effectively reduces pressure, the system will react by shifting the equilibrium to the side with more moles of gas to increase the pressure back up. Simply put, it's like a balance scale; when you add weight to one side (change conditions), the scale will tip (shift equilibrium) to bring itself back into balance.
Equilibrium Shift
An 'equilibrium shift' refers to the movement of a reversible reaction's equilibrium position in response to a change in conditions. The shift indicates which direction the reaction moves to re-establish equilibrium. In the context of increasing the reaction volume, a shift in equilibrium typically occurs towards the side with more moles of gaseous reactants or products. This is because increasing the volume decreases the concentration of gases and, by Le Chatelier's principle, the system will shift to increase the concentration again.

Using the textbook exercises, students can apply this concept to various reactions for prediction. For example, in reaction (b), increasing volume causes the equilibrium to shift to the right where there are more gas molecules. This is a visual way to see chemical systems 'respond' to the stress of volume change, which brings a dynamic understanding of chemical reactions beyond static memorization.
Reaction Volume Change
The impact of reaction volume changes on the equilibrium can be understood through the lens of the ideal gas law, which states that pressure and volume are inversely proportional for a gas at a constant temperature (PV = nRT). When the volume of a reaction container is increased, the pressure of the gases inside decreases, assuming the number of moles (n) and temperature (T) are constant.

This decrease in pressure is what influences the equilibrium shift, as the system compensates for the drop by shifting towards the side with more moles of gas. From the exercise given, we can see practical examples of this relationship. For instance, in reaction (a) the system shifts towards the side with 4 moles of gas— the reactants — on increasing volume. Understanding how volume changes affect pressure and, in turn, influence equilibrium, is critical to mastering the concepts of chemical equilibrium in real-world scenarios.

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Most popular questions from this chapter

An equilibrium mixture contains \(0.60 \mathrm{~g}\) solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of \(2.60\) atm and \(2.89\) atm, respectively. Calculate the value of \(K_{\mathrm{e}}\) for the reaction \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\)

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following cases. a. \(1.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) and \(2.0 \mathrm{~g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a \(1.0-\mathrm{L}\) flask. b. \(1.0\) mole of pure \(\mathrm{HOCl}\) is placed in a \(2.0-\mathrm{L}\) flask.

For the reaction $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3}\) \(M,\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{~N}_{2}(g)\right]=5.3 \times 10^{-2} M\), and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

The gas arsine, \(\mathrm{AsH}_{3}\), decomposes as follows: $$ 2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g) $$ In an experiment at a certain temperature, pure \(\operatorname{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of \(392.0\) torr. After 48 hours the pressure in the flask was observed to be constant at \(488.0\) torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(\mathrm{~g})\). b. Calculate \(K_{\mathrm{p}}\) for this reaction.

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