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Pure phosgene gas \(\left(\mathrm{COCl}_{2}\right), 3.00 \times 10^{-2} \mathrm{~mol}\), was placed in a 1.50 - \(\mathrm{L}\) container. It was heated to \(800 \mathrm{~K}\), and at equilibrium the pressure of \(\mathrm{CO}\) was found to be 0.497 atm. Calculate the equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$

Short Answer

Expert verified
The equilibrium constant KP for the reaction is found to be 2.42.

Step by step solution

01

Find the initial partial pressure of COCl2

First, calculate the initial pressure of the COCl2 gas using the ideal gas law: P = n/V where n is the number of moles and V is the volume (in this case, 3.00 x 10-2 mol and 1.50 L respectively). This value will be useful to find the amount of COCl2 that decomposes.
02

Calculate the change in moles

Find the number of moles of CO and Cl2 formed, using the pressure of CO at equilibrium (0.497 atm) and the ideal gas law rearranged as n = PV/RT (where R is the ideal gas constant and T is the absolute temperature). This will also give us the amount of COCl2 that has decomposed.
03

Equilibrium pressures for all gases

The COCl2 decomposes into CO and Cl2. So, subtract the amount of COCl2 that decomposes from the initial amount to get the amount at equilibrium. Then, use the ideal gas law to convert these amounts to pressures.
04

Find the equilibrium constant KP

Substitute these equilibrium pressures into the expression for the equilibrium constant KP = (PCO x PCl2) / PCOCl2.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is central to grasping the behavior of reactions in a closed system. It's a state where the rate of the forward reaction equals the rate of the reverse reaction, resulting in a stable, observable condition. In this state, the concentrations (or pressures in the case of gases) of reactants and products remain constant over time, though not necessarily equal.

In the context of the phosgene gas exercise, chemical equilibrium implies that despite the ongoing interconversion between phosgene, carbon monoxide (CO), and chlorine gas (Cl2), their amounts in the container, when expressed as partial pressures, do not change once equilibrium is established.
Ideal Gas Law
The ideal gas law is a fundamental equation that describes the relationship between the pressure (P), volume (V), temperature (T), and the number of moles (n) of an ideal gas. Expressed as PV = nRT, where R is the universal gas constant, this law allows us to calculate any one of the variables if the other three are known.

Using this law, we can elucidate the conditions of our phosgene gas before and after it reaches equilibrium. For instance, initially we can calculate the pressure of COCl2 before it dissociates. At equilibrium, we apply the ideal gas law to find the pressure changes and thus the quantities of CO and Cl2 formed.
Le Chatelier's Principle
Le Chatelier's principle is a qualitative tool that predicts how a system at equilibrium will respond to external changes, such as variations in pressure, temperature, or concentration. Essentially, it posits that if an external stress is applied, the system will adjust itself in a direction to counteract that stress and re-establish equilibrium.

In the scenario of phosgene gas being heated to a high temperature, the equilibrium position is expected to shift to either produce more products or revert to more reactants, depending on which direction alleviates the effect of the applied heat. Understanding this principle helps in predicting and calculating the equilibrium positions under varying conditions.

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Most popular questions from this chapter

Consider the reaction $$ \begin{aligned} 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) & \\ \Delta H^{\circ}=&-198.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to \((\mathrm{a})\) increase the temperature, (b) increase the pressure, (c) increase \(\mathrm{SO}_{2},\) (d) add a catalyst, (e) add helium at constant volume.

A 2.50 -mol quantity of \(\mathrm{NOCl}\) was initially placed in a 1.50-L reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the NOCl had dissociated: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium constant \(K_{c}\) for the reaction.

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ is 54.3 at \(430^{\circ} \mathrm{C}\). At the start of the reaction there are 0.714 mole of \(\mathrm{H}_{2}, 0.984\) mole of \(\mathrm{I}_{2}\), and 0.886 mole of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium.

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$ \mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g) $$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

Consider the following equilibrium process at \(700^{\circ} \mathrm{C}\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) $$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}, 1.35 \times\) \(10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0-L flask at equilibrium. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

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