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Chapter 15: Problem 44

Heating solid sodium bicarbonate in a closed vessel established this equilibrium: \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system, (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system, (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.

Short Answer

Expert verified
(a) The equilibrium will shift to the right. (b) Adding \(Na_{2}CO_{3}\) will not affect the equilibrium position. (c) Removing \(NaHCO_{3}\) will not affect the equilibrium position.

Step by step solution

01

Effect of Removing \(CO_{2}\) from the system

According to Le Chatelier's Principle, if \(CO_{2}\) is removed from the system, the reaction will shift to the right to replace the quantity of \(CO_{2}\) that was removed. This is because the system will want to achieve its equilibrium state again by increasing the concentration of \(CO_{2}\) to its original level. This means the reaction will lean towards the products.
02

Effect of Adding \(Na_{2}CO_{3}\) to the system

Adding more of \(Na_{2}CO_{3}\) to the system does not affect the equilibrium position as it is a solid. In terms of equilibrium, the concentration of a solid or liquid does not change as it doesn’t have a volume in which to be spread.
03

Effect of Removing \(NaHCO_{3}\) from the system

If some of the solid \(NaHCO_{3}\) were removed from the system, it would not affect the equilibrium position. This is because in the equilibrium expression, solids are not considered as their concentrations do not change.

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Most popular questions from this chapter

When heated, ammonium carbamate decomposes as $$ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ At a certain temperature the equilibrium pressure of the system is 0.318 atm. Calculate \(K_{P}\) for the reaction.

The "boat" form and "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconverts as shown here: In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair \(\longrightarrow\) boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol}\). If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1}\), what is \(k_{1}\) at \(298 \mathrm{~K}\) ? The equilibrium constant \(K_{c}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, andit is also responsible for the acid rain phenomenon. The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment 2.00 moles of \(\mathrm{SO}_{2}\) and 2.00 moles of \(\mathrm{O}_{2}\) were initially present in a flask. What must the total pressure be at equilibrium to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

What is meant by "the position of an equilibrium"? Does the addition of a catalyst have any effects on the position of an equilibrium?
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