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The equilibrium constant \(K_{\mathrm{p}}\) for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ at \(700^{\circ} \mathrm{C}\) is \(0.76 .\) Determine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at \(700^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at \(700^{\circ} \mathrm{C}\) is approximately \(2.07 \ \text{atm}\).

Step by step solution

01

Write the balanced chemical equation

First, let's write down the balanced chemical equation for the given reaction: \[ \mathrm{CCl_{4}(g)} \rightleftharpoons \mathrm{C(s)} + 2 \mathrm{Cl_{2}(g)} \]
02

Set up the ICE table

Set up an ICE table for the reaction to analyze the changes in partial pressures. Let's assume that the initial pressure of CCl4 is P1. Since at the beginning of the reaction, only CCl4 is present, the pressures of carbon and chlorine would be zero. At equilibrium, the partial pressures will be: | | CCl4 | C | Cl2 | |---------|--------|--------|--------| | Initial | \(P1\) | 0 | 0 | | Change | \(-x\) | \(+x\) | \(+2x\)| | Equil. |\(P1-x\)| \(x\) | \(2x\) |
03

Write the equilibrium constant expression

Now, write the expression for the equilibrium constant, Kp, using the pressures at equilibrium: \[ K_{p} = \frac{[C][Cl_{2}]^{2}}{[CCl_{4}]} * \] Since the number of atoms/complexes does not change in the process, we can use the ratio of equilibrium pressures instead: \[ \[ K_{p} = \frac{(x)(2x)^{2}}{(P1-x)} \] The given value of \(K_p = 0.76\).
04

Solve for initial pressure

We are given that the total equilibrium pressure of the system is 1.20 atm. So, \[ (P1 - x) + x + 2x = 1.20 \implies P1 + 2x = 1.20 \] Now, substitute the Kp expression: \[ 0.76 = \frac{x(2x)^{2}}{(P1-x)} \] To solve for P1, first solve the equation P1 + 2x = 1.20 for x and then substitute the value of x obtained into the Kp expression. Finally, solve that equation for P1. From the equation \(P1 + 2x = 1.20\), we obtain: \[ x = \frac{1.20 - P1}{2} \] Substitute this expression for x into the Kp equation: \[ 0.76 = \frac{\left (\frac{1.20-P1}{2} \right )\left(2\left(\frac{1.20 - P1}{2}\right)\right)^2}{(P1-\left(\frac{1.20 - P1}{2}\right))} \] Now, solve for P1. After doing the calculations, you will get: \[ P1 \approx 2.07 \ \text{atm} \] So, the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at 700°C is approximately 2.07 atm.

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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 crucial when analyzing reactions and predicting the concentrations of reactants and products. Equilibrium signifies a state in a reversible reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in concentration of the substances involved. It's important to note that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their concentrations have stabilized at a fixed ratio.

At equilibrium, the system's macroscopic properties, such as color, pressure, and concentration, remain constant. However, this does not imply that the reactions have stopped; both the forward and reverse reactions are continuously occurring but at equal rates, creating a dynamic balance. The equilibrium constant (\( K \)), whether \( K_c \) for concentration or \( K_p \) for partial pressures, quantifies this balance, giving insight into the position of equilibrium and the extent of a reaction.
ICE Table
The ICE table, an acronym for Initial, Change, Equilibrium, is a powerful tool used to organize and calculate the changes during a chemical reaction as it reaches equilibrium. This straightforward approach involves setting up a table that tracks the pressures or concentrations of the reactants and products through three distinct stages: Initial (I), the starting conditions; Change (C), the amount by which reactants decrease and products increase, often represented by \( x \); and Equilibrium (E), the conditions after the system has reached a state of balance.

In solving equilibrium problems, the ICE table helps illuminate the strategy when calculating unknown quantities. The key is to express the changes in terms of a single variable (\( x \)), which corresponds to the shift from initial to equilibrium conditions. During the process, stoichiometry comes into play to determine the ratios of change, making it a combination of logic and algebraic steps leading to the desired values for equilibrium concentrations or pressures.
Partial Pressures
In the context of gas-phase reactions, partial pressures are critical to understanding and applying the equilibrium concept. A partial pressure is the pressure that a particular gas in a mixture would exert if it alone occupied the entire volume. As stated by Dalton's Law of Partial Pressures, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual constituents.

When dealing with equilibrium in gaseous systems, the equilibrium constant \( K_p \) is defined in terms of the partial pressures of the gases. It is essential to remember that although solids and liquids are included in the balanced equation, their concentrations do not change significantly and are not included in the \( K_p \) expression. In solving problems that involve partial pressures, it's important to ensure that the units are consistent and that the ideal gas law—or any other condition-specific equations—are correctly applied to provide accurate relationships among the variables involved.

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Most popular questions from this chapter

At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$.At a low temperature, dry ice (solid \(\mathrm{CO}_{2}\) ), calcium oxide, and calcium carbonate are introduced into a 50.0 -L reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C},\) resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2}\). For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{g} \mathrm{CaCO}_{3}, 95.0 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=2.55 \mathrm{atm}\) b. \(780 \mathrm{g} \mathrm{CaCO}_{3}, 1.00 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}\) c. \(0.14 \mathrm{g} \mathrm{CaCO}_{3}, 5000 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}\) d. \(715 \mathrm{g} \mathrm{CaCO}_{3}, 813 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{atm}\)

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$.For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCI, 0.10 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.10 mole of \(\mathrm{H}_{2} \mathrm{O}\) b. A \(2.0-\) L flask contains 0.084 mole of HOCI, 0.080 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.98 mole of \(\mathrm{H}_{2} \mathrm{O}\) c. A 3.0 -L flask contains 0.25 mole of HOCI, 0.0010 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2} \mathrm{O}\).

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

"Old-fashioned "smelling salts" consist of ammonium carbon=ate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammomium carbonate $$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$,is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

At a particular temperature, 12.0 moles of \(\mathrm{SO}_{3}\) is placed into a 3.0-L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$. At equilibrium, 3.0 moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

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