Question

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 $\mathrm{atm}$ at ${700}^{\circ }\mathrm{C}.$

Short answer

The initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at 700°C is approximately 0.9696 atm.

Step-by-step solution

Initial and Equilibrium Pressures

Let x be the moles of CCl₄ that reacted at equilibrium. The initial pressure of CCl₄ is P(CCl₄) and since there was no Cl₂ in the beginning, we can write the pressures at equilibrium as: P(CCl₄) - x for CCl₄ 2x for Cl₂ Note that, since Carbon in the reaction is in solid state, it doesn't affect the pressures. The total equilibrium pressure is given as 1.20 atm. So, (P(CCl₄) - x) + 2x = 1.20 Simplifying the equation, we get: P(CCl₄) + x = 1.20

Write the expression for Kp

We are given the equilibrium constant Kp = 0.76. The expression for Kp, using the pressures of gases at equilibrium, is: Kp = ${\displaystyle \frac{(P(C{l}_2){)}^2}{P(CC{l}_4)}}$ At equilibrium, the pressure of Cl₂ is 2x. Thus, we can rewrite the expression as: 0.76 = ${\displaystyle \frac{(2x{)}^2}{P(CC{l}_4)-x}}$

Solving for the initial pressure of CCl₄

Now, we can rewrite our Kp equation in terms of x and P(CCl₄) only: 0.76 = ${\displaystyle \frac{4x^2}{P(CC{l}_4)-x}}$ Using the equation from Step 1 (P(CCl₄) + x = 1.20), we can eliminate P(CCl₄) and solve for x: P(CCl₄) - x = 1.20 - 2x Now, we can substitute this expression in our Kp equation: 0.76 = ${\displaystyle \frac{4x^2}{1.20-2x}}$ Let's isolate x in the equation: 0.76(1.20 - 2x) = 4x² After simplifying the equation, we get: 0.912 - 1.52x = 4x² Rearranging terms, we obtain the quadratic equation: 4x² + 1.52x - 0.912 = 0 We can solve this quadratic equation for x using any method (e.g., quadratic formula, factoring, etc.). Here, we will use the quadratic formula: x = ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ where a = 4, b = 1.52, and c = -0.912. Calculating x, we get two possible values: x₁ ≈ 0.2304 and x₂ ≈ -0.991. Since x represents moles reacting and cannot be negative, we discard x₂ and use x₁: x ≈ 0.2304

Calculate the initial pressure of CCl₄

Now that we have the value of x, we can find the initial pressure of CCl₄ using the equation from Step 1: P(CCl₄) + x = 1.20 P(CCl₄) + 0.2304 = 1.20 P(CCl₄) ≈ 1.20 - 0.2304 P(CCl₄) ≈ 0.9696 atm The initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at 700°C is approximately 0.9696 atm.

Key concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted as $K_p$, is a crucial concept in understanding how chemical reactions proceed and reach equilibrium, especially in gaseous systems. It provides insights into the ratio of product concentrations to reactant concentrations at a given temperature.

For reactions involving gases, $K_p$ is expressed in terms of partial pressures rather than concentrations. This transformation is especially helpful because gases can expand to fill their containers, making their behavior more variable based on pressure.

For the reaction ${\mathrm{CCl}}_4(g)\rightleftharpoons \mathrm{C}(s)+2{\mathrm{Cl}}_2(g)$, the equilibrium constant is given by the equation: $$K_p=\frac{(P({\mathrm{Cl}}_2){)}^2}{P({\mathrm{CCl}}_4)}$$ Here, $P({\mathrm{Cl}}_2)$ and $P({\mathrm{CCl}}_4)$ denote the equilibrium partial pressures of chlorine and carbon tetrachloride, respectively. It's important to remember solid substances, like carbon in this case, are not included in equilibrium constant expressions.

Carbon Tetrachloride

Carbon tetrachloride (${\mathrm{CCl}}_4$) is a volatile chemical compound often used in different industrial applications, such as a solvent for fats and oils. In the context of chemical reactions, ${\mathrm{CCl}}_4$ behaves as a gaseous reactant in equilibrium studies, where its ability to change phase rapidly is taken into account.

During chemical reactions, ${\mathrm{CCl}}_4$ can decompose under certain conditions, such as high temperatures, to release chlorine gas $({\mathrm{Cl}}_2)$. In the given exercise, this decomposition happens at 700°C, forming a dynamic equilibrium with chlorine gas. As you understand equilibrium scenarios, note that the initial pressure of ${\mathrm{CCl}}_4$ is a key factor in determining how the reaction will proceed and stabilize.

Knowing this allows us to better predict the behavior of the system and understand how pressure changes due to reactions.

Quadratic Equation

Solving equilibrium problems often involves mathematical equations, and in this exercise, we encounter a quadratic equation. This emerges when rearranging and equating expressions derived from the equilibrium setup.

The quadratic equation in standard form is $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, and $x$ represents the variable we seek to determine. In our exercise, the decomposition and pressure relationships of ${\mathrm{CCl}}_4$ and ${\mathrm{Cl}}_2$ leads us to form the equation: $$4x^2+1.52x-0.912=0$$ To solve for $x$, we employ the quadratic formula: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ This formula helps us find the numerical value of $x$, representing the extent of decomposition, by providing roots of the equation. Only positive solutions are considered, as negative amounts of substances are physically unrealizable.

Chemical Equilibrium

Chemical equilibrium arises in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant over time.

In the scenario described by the equation ${\mathrm{CCl}}_4(g)\rightleftharpoons \mathrm{C}(s)+2{\mathrm{Cl}}_2(g)$, equilibrium explains why despite the reaction proceeding to form ${\mathrm{Cl}}_2$, carbon tetrachloride remains in the system. The system ceases to change in its macroscopic properties, but reactions still occur at a molecular level.

Understanding equilibrium helps us predict whether, at any given point, more reactants will convert into products or vice versa. It's a state of balance that allows chemists to figure out the conditions needed to either maximize yield or maintain stability, vital for industrial and laboratory processes.