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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 \(0.76\) atm. Determine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of \(1.20 \mathrm{~atm}\) 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°C is approximately \( 1.027 \, atm \).

Step by step solution

01

Write the balanced chemical equation and the expression for Kp

Here is the balanced chemical equation and the Kp expression: \( CCl_4(g) \rightleftharpoons C(s) + 2Cl_2(g) \) \[ K_p = \frac{[Cl_2]^2}{[CCl_4]} \]
02

Setup the ICE table

To set up the ICE table, we can represent the initial pressure of CCl4 as P and all the other components as 0 since we don't have any information about them. | | CCl4 | C | Cl2 | |----------|-------|---------|--------| | Initial | P | 0 | 0 | | Change | -x | +x | +2x | | Equilbr. | P - x | x | 2x | Since carbon solid is not involved in the Kp expression, it won't affect the table.
03

Substitute the equilibrium values from the ICE table into the Kp expression

Substitute the equilibrium values from the ICE table into the Kp expression: \[ K_p = \frac{(2x)^2}{(P-x)} \]
04

Use the given Kp value to solve for x

Since Kp = 0.76 at 700°C, we substitute this into the equation from Step 3 and solve for x: \[ 0.76 = \frac{(2x)^2}{(P-x)} \]
05

Solve for the total pressure at equilibrium

The total equilibrium pressure is given as 1.20 atm. This includes the pressure of the CCl4 and the Cl2 at equilibrium: \[ P_{total} = (P-x) + 2x = 1.20 \]
06

Solve the system of equations to find the initial pressure of CCl4

We have a system of two equations with two variables (P and x): 1) \( 0.76 = \frac{(2x)^2}{(P-x)} \) 2) \( 1.20 = (P-x) + 2x \) First, solve Eq. (2) for P: \( P = 1.20 - x \) Then, substitute this into Eq. (1) and solve for x: \[ 0.76 = \frac{(2x)^2}{(1.20 - x) - x} \] Solve for x, approximately 0.173 atm and substitute this value back into the equation for P above: \( P = 1.20 - 0.173 \)
07

Calculate the initial pressure of CCl4

Finally, we can calculate the initial pressure of CCl4: \( P = 1.027 \, atm \) Thus, the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at 700°C is approximately 1.027 atm.

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Key Concepts

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

Equilibrium Constant Kp
The equilibrium constant, symbolized as \( K_p \), is a vital concept in understanding chemical reactions that occur in a gaseous state. It represents the ratio of the pressure of the products to the reactants at equilibrium, each raised to the power of their stoichiometric coefficients. For the reaction
  • \( \mathrm{CCl}_{4}(g) \leftrightharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) \)
\( K_p \) can be expressed as:\[ K_p = \frac{[\mathrm{Cl}_2]^2}{[\mathrm{CCl}_4]} \]Here, the concentration of carbon, \( C(s) \), is omitted because it is a pure solid and does not affect the equilibrium position in the reaction quotient. By understanding \( K_p \), chemists can predict the extent of a reaction and the conditions under which the system reaches equilibrium at a given temperature.
Carbon Tetrachloride
Carbon tetrachloride (\( \mathrm{CCl}_4 \)) is a molecule that consists of one carbon atom surrounded by four chlorine atoms, making it a tetrahedral compound. It is a colorless liquid with various industrial applications, but it is most relevant in this context for being the reactant in our chemical equilibrium problem.In this reaction, it is decomposed into carbon and chlorine gas under high temperatures. The pressure of \( \mathrm{CCl}_4 \) at equilibrium depends on the initial amount introduced into the system. By manipulating the pressure, chemists can control the equilibrium position and stabilize the production of desired products. This makes \( \mathrm{CCl}_4 \) not only significant from a theoretical perspective but also useful in practical applications and experiments.
ICE Table
An ICE table, which stands for Initial, Change, Equilibrium, is a powerful tool used to track the changes in concentration or pressure during a chemical reaction. It helps in organizing the initial values, changes that occur, and equilibrium concentrations of all species involved in a chemical reaction.For the equilibrium reaction:
  • \( \mathrm{CCl}_{4}(g) \leftrightharpoons \mathrm{C}(s) + 2 \mathrm{Cl}_{2}(g) \)
at 700°C, the ICE table is structured as follows:
  • Initial pressures: \( \mathrm{CCl}_4 \) is \( P \), whereas \( \mathrm{C} \) and \( \mathrm{Cl}_2 \) both start at zero.
  • Change in pressures: \( \mathrm{CCl}_4 \) decreases by \( x \), \( \mathrm{C} \) increases by \( x \), and \( \mathrm{Cl}_2 \) increases by \( 2x \).
  • Equilibrium pressures: \( \mathrm{CCl}_4 \) ends up as \( P-x \), \( \mathrm{C} \) as \( x \), and \( \mathrm{Cl}_2 \) as \( 2x \).
By substituting the equilibrium values into expressions derived from \( K_p \), one can solve for unknown values such as the initial pressure of reactants.
Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balance occurs when the concentrations of reactants and products remain constant over time, although they may not necessarily be equal.
  • In the context of the equation \( \mathrm{CCl}_{4}(g) \leftrightharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) \), reaching equilibrium means the rate at which \( \mathrm{CCl}_{4} \) decomposes into \( \mathrm{C} \) and \( \mathrm{Cl}_2 \) is equalized by the rate at which \( \mathrm{Cl}_2 \) combines to form \( \mathrm{CCl}_{4} \).
  • The conditions which determine equilibrium include temperature, pressure, and the intrinsic properties of the reactants and products.
Understanding chemical equilibrium allows chemists to predict how a reaction will respond to external changes, such as a change in pressure or temperature, and maintain control over reaction conditions to achieve favorable outcomes.

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Most popular questions from this chapter

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\) b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) d. \(2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\)

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a \(0.256\) -mol sample of naphthalene has a mass of \(32.8 \mathrm{~g}\). What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene \((s) \rightleftharpoons\) naphthalene \((g)\) \(K=4.29 \times 10^{-6}(\) at \(298 \mathrm{~K})\) If \(3.00 \mathrm{~g}\) solid naphthalene is placed into an enclosed space with a volume of \(5.00 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\), what percentage of the naphthalene will have sublimed once equilibrium has been established?

Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium: $$ \underset{\text { Purple }}{\mathrm{CoCl}_{2}(s)}+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \underset{\text { Pink }}{\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)} $$ What color will such an indicator be if rain is imminent?

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\ \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\ \mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\ \mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14} \end{array} $$ determine the values for the equilibrium constants for the following reactions. a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

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