Write the expression of the reaction quotient for the ionization of HOCN in water.

Short Answer

Expert verified
The reaction quotient for the ionization of HOCN in water is given by Q = [OCN^-][H_3O^+]/[HOCN].

Step by step solution

01

Write down the balanced chemical equation for the ionization of HOCN in water

First, write the balanced chemical equation for the ionization of HOCN (cyanic acid) in water. The reaction is as follows: HOCN + H_2O ⇌ OCN^- + H_3O^+.
02

Write the expression for the reaction quotient (Q)

Next, use the balanced chemical equation to write the expression for the reaction quotient, Q. The formula is Q = [products]/[reactants], with each concentration raised to the power of its coefficient in the balanced equation. Since each reactant and product has a coefficient of 1, Q is given by: Q = [OCN^-][H_3O^+]/[HOCN].

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Key Concepts

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

Ionization of HOCN
Understanding the ionization of HOCN (cyanic acid) is essential for grasping core principles in chemistry, particularly acid-base reactions. Ionization refers to the process where a molecule, such as HOCN, reacts with water and dissociates into ions. In this particular reaction, HOCN donates a proton (H+) to water, yielding the cyanate ion (OCN-) and the hydronium ion (H3O+).

Ionization is usually represented with a chemical equation: \[ \text{HOCN} + \text{H}_2\text{O} \rightleftharpoons \text{OCN}^- + \text{H}_3\text{O}^+ \].
This reversible reaction showcases the dynamic nature of ionization, which can go forwards or backwards depending on various conditions such as concentration and temperature. In this context, the extent of ionization is quantitatively expressed by the reaction quotient.
Chemical Equilibrium
Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of the reactants and products over time. It's a balance of transfer rates, not necessarily equal concentrations of reactants and products. For the ionization of HOCN in water, equilibrium can be established after certain time, where the reaction reaches a state where the formation of OCN- and H3O+ balances out with the recombination of these ions to form HOCN.

At equilibrium, the reaction quotient (Q) becomes equal to the equilibrium constant (K), and thus Q can be used to discern the direction of the reaction. A Q value smaller than K implies that the forward reaction is favored, while a Q value larger than K indicates that the reverse reaction is favored to achieve equilibrium.
Chemical Reaction in Water
A chemical reaction in water, often involving ionization, is central to many processes in chemistry, such as dissolution, neutralization, and redox reactions. Water is known as the 'universal solvent' due to its ability to dissolve many substances, and its role in facilitating chemical reactions is pivotal. In the ionization of HOCN, water acts as a solvent and also as a reactant.

The process of ionization in water is influenced by water's polarity, which allows it to stabilize ions, making it an ideal medium for the reaction to occur. When writing a reaction quotient for a chemical reaction in water, it's important to remember that the concentration of water is typically not included in the expression because it is present in large excess and its concentration changes negligibly during the reaction.

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Most popular questions from this chapter

The equilibrium constant \(\left(K_{c}\right)\) for this reaction is 5.0 at a given temperature. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) (a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of \(\mathrm{CO}, 0.30\) mol of water vapor, and \(0.90 \mathrm{mol}\) of \(\mathrm{H}_{2}\) in a liter. How many moles of \(\mathrm{CO}_{2}\) were there in the equilibrium mixture? (b) Maintaining the same temperature, additional \(\mathrm{H}_{2}\) was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of \(\mathrm{CO}, 0.30\) mol of water vapor, and 1.2 mol of \(\mathrm{H}_{2}\) in a liter. How many moles of \(\mathrm{CO}_{2}\) were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.

Write the mathematical expression for the reaction quotient, \(Q_{c}\), for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)=2 \mathrm{NO}_{2}(g)\) (d) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{NH}_{4} \mathrm{Cl}(s)=\mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) (f) \(2 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightleftharpoons 2 \mathrm{PbO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) (g) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)=2 \mathrm{H}_{2} \mathrm{O}(l)\) (h) \(S_{8}(g) \rightleftharpoons 8 S(g)\)

Sodium sulfate 10-hydrate, \(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O},\) dehydrates according to the equation \(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+10 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{P}=4.08 \times 10^{-25} \mathrm{at} 25^{\circ} \mathrm{C}\) What is the pressure of water vapor at equilibrium with a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{NaSO}_{4} ?\)

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, \(S O_{3}\), from sulfur dioxide, \(S O_{2},\) and oxygen, \(O_{2}\), shown here. At high temperatures, the rate of formation of \(S O_{3}\) is higher, but the equilibrium amount (concentration or partial pressure) of \(\mathrm{SO}_{3}\) is lower than it would be at lower temperatures. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (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?

Calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for each of the following reactions from the value of \(\Delta G^{\circ}\) given. (a) \(\mathrm{I}_{2}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{ICl}(g) \quad \Delta G^{\circ}=-10.88 \mathrm{kJ}\) (b) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=3.4 \mathrm{kJ}\) (c) \(\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g) \quad \Delta G^{\circ}=-39 \mathrm{kJ}\) (d) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad \Delta G^{\circ}=-141.82 \mathrm{kJ}\) (e) \(\mathrm{CS}_{2}(g) \longrightarrow \mathrm{CS}_{2}(l) \quad \Delta G^{\circ}=-1.88 \mathrm{kJ}\)

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