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Phosgene, \(\mathrm{COCl}_{2}(g)\), a toxic gas used in the synthesis of a variety of organic compounds, decomposes according to $$ \mathrm{COCl}_{2}(g) \leftrightharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ A sample of phosgene gas at an initial concentration of \(0.500 \mathrm{M}\) is heated at \(527^{\circ} \mathrm{C}\) in a reaction vessel. At equilibrium, the concentration of \(\mathrm{CO}(g)\) was found to be \(0.046 \mathrm{M}\). Calculate the equilibrium constant for the reaction equation at \(527^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The equilibrium constant \( K_c \) is approximately 0.00466.

Step by step solution

01

Determine the Change in Concentrations

First, notice the balanced chemical equation: \( \mathrm{COCl}_2(g) \leftrightharpoons \mathrm{CO}(g) + \mathrm{Cl}_2(g) \). Initially, only \( \mathrm{COCl}_2 \) was present with a concentration of 0.500 M. At equilibrium, the concentration of \( \mathrm{CO} \) is given as 0.046 M. This means the change in concentration for \( \mathrm{CO} \) is 0.046 M. Since the stoichiometry is 1:1:1, the change for \( \mathrm{Cl}_2 \) is also 0.046 M, and for \( \mathrm{COCl}_2 \), it is decreased by 0.046 M.
02

Calculate Equilibrium Concentrations

Knowing the initial concentration of \( \mathrm{COCl}_2 \) was 0.500 M and it decomposed by 0.046 M, the equilibrium concentration of \( \mathrm{COCl}_2 \) is: \( 0.500 - 0.046 = 0.454 \mathrm{M} \). As \( \mathrm{CO} \) and \( \mathrm{Cl}_2 \) both started at 0 M and gained 0.046 M, their equilibrium concentrations are both 0.046 M.
03

Set Up the Equilibrium Expression

The equilibrium expression for the reaction is: \( K_c = \frac{[\mathrm{CO}][\mathrm{Cl}_2]}{[\mathrm{COCl}_2]} \). Plug in the equilibrium concentrations obtained: \( [\mathrm{CO}] = 0.046 \mathrm{M} \), \([\mathrm{Cl}_2] = 0.046 \mathrm{M}\), and \([\mathrm{COCl}_2] = 0.454 \mathrm{M}\).
04

Calculate the Equilibrium Constant

Substitute the values into the equilibrium expression: \[ K_c = \frac{(0.046)(0.046)}{0.454} \]. This calculates to be \( K_c = \frac{0.002116}{0.454} \approx 0.00466 \). Therefore, the equilibrium constant \( K_c \) at \( 527^\circ C \) is approximately 0.00466.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to a state in a chemical reaction where the rate at which the reactants convert into products equals the rate at which the products convert back into reactants. In the case of the phosgene gas decomposition, the reaction achieves equilibrium once the concentrations of CO, Cleightsub>2, and COCleightsub>2 no longer change over time.
A key feature of chemical equilibrium is that it is dynamic. This means although the concentrations remain constant, the molecules continue to react with each other. The forward and reverse reactions occur simultaneously and at the same rate.
At equilibrium, the system is governed by the equilibrium constant, \( K_c \), which provides critical insight into the relative concentrations of reactants and products. Understanding this balance helps us predict how a system responds to changes in conditions.
Reaction Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's essential to understand reaction stoichiometry to calculate changes in concentrations as a reaction approaches equilibrium.
In our example, the decomposition of phosgene ( COCl eightsub>2) into carbon monoxide (CO) and chlorine ( Cl eightsub>2) follows a 1:1:1 stoichiometric ratio. This implies that every molecule of phosgene decomposes to form one molecule of CO and one molecule of Cl eightsub>2.
This 1:1:1 ratio means that any change in the concentration of one component in the reaction mixture results in an equivalent change in the concentrations of the others. For example, a decrease of 0.046 M in phosgene results in an increase of 0.046 M each for carbon monoxide and chlorine.
Concentration Changes
In the context of a chemical reaction approaching equilibrium, changes in concentration are pivotal. We begin with initial concentrations and need to calculate how these change over time until equilibrium is established.
For the decomposition of phosgene, the initial concentration of COCl eightsub>2 is given as 0.500 M, while both CO and Cl eightsub>2 start at 0 M. As the reaction proceeds, phosgene decreases by 0.046 M, and simultaneously, the concentrations of CO and Cl eightsub>2 increase by 0.046 M each.
It is crucial to monitor these changes, as they allow us to calculate the equilibrium concentrations of each component. This information is then used to determine the equilibrium constant. Understanding these shifts helps in predicting how the system will react if conditions change.
Equilibrium Expression
The equilibrium expression is a mathematical equation that uses the concentrations of reactants and products to calculate the equilibrium constant, \( K_c \). This expression depends on the balanced chemical equation and the stoichiometry.
For the reaction \( \mathrm{COCl}_2(g) \leftrightharpoons \mathrm{CO}(g) + \mathrm{Cl}_2(g) \), the equilibrium expression is \[ K_c = \frac{[\mathrm{CO}][\mathrm{Cl}_2]}{[\mathrm{COCl}_2]} \].
Plugging in the equilibrium concentrations yields: \([\mathrm{CO}] = 0.046 \mathrm{M}\), \([\mathrm{Cl}_2] = 0.046 \mathrm{M}\), and \([\mathrm{COCl}_2] = 0.454 \mathrm{M}\).
This gives us the equation: \[ K_c = \frac{(0.046)(0.046)}{0.454} \approx 0.00466 \].
Understanding and setting up the equilibrium expression is critical for calculating \( K_c \), and it provides insights into the extent to which reactants are converted into products in a reaction.

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Most popular questions from this chapter

The decomposition of ammonium hydrogen sulfide is an endothermic reaction. The equation for the reaction is $$ \mathrm{NH}_{4} \mathrm{HS}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ A \(5.260\) -gram sample of solid ammonium hydrogen sulfide is placed in an evacuated \(9.00\) -liter container at \(25^{\circ} \mathrm{C}\). After equilibrium is established, the total pressure inside the vessel is \(0.659 \mathrm{~atm}\). Some solid ammonium hydrogen sulfide remains in the flask. (a) What is the value of the equilibrium constant, \(K_{\mathrm{p}} ?\) (b) What percentage of the solid placed in the flask has reacted?

Consider the methanation reaction described by the chemical equation $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ It was found that \(0.618\) moles of \(\mathrm{CO}(g), 0.387\) moles of \(\mathrm{H}_{2}(\mathrm{~g}), 0.387 \mathrm{moles}\) of \(\mathrm{CH}_{4}(g)\), and \(0.387\) moles of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) were present in an equilibrium mixture in a 1.00-liter container. All the water vapor was removed and the system allowed to come to equilibrium again. Calculate the concentration of all gases in the new equilibrium system. (You must solve the resulting equation by trial and error.)

According to Table 19.1, \(K_{c}=0.20 \mathrm{M}\) at \(100^{\circ} \mathrm{C}\) for the chemical equation $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \leftrightharpoons 2 \mathrm{NO}_{2}(g) $$ Calculate \(K_{\mathrm{p}}\) at the same temperature in units of bars.

Given that (1) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(g) \leftrightharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) \(K_{\mathrm{P}_{1}}=1.44\) (2) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \leftrightharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)\) \(K_{\mathrm{P}_{2}}=25.6 \mathrm{~atm}^{2}\) calculate the value of \(K_{p}\) for the equation (3) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \leftrightharpoons \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g)\)

Sodium hydrogen carbonate, commonly called sodium bicarbonate, is used in baking soda and in fire extinguishers as a source of \(\mathrm{CO}_{2}(g) .\) It decomposes according to the equation $$ 2 \mathrm{NaHCO}_{3}(s) \leftrightharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Given that \(K_{\mathrm{p}}=0.26 \mathrm{bar}^{2}\) at \(125^{\circ} \mathrm{C}\), calculate the partial pressures of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) in units of bars at equilibrium when \(\mathrm{NaHCO}_{3}(s)\) is heated to \(125^{\circ} \mathrm{C}\) in a closed vessel.

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