Question

At 700 K, CCl4 decomposes to carbon and chlorine. The Kp value for the decomposition is 0.76. Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.9 atm at equilibrium.

Answer 1

Answer: The starting pressure of $CCl_4$ is 1.39 atm.

Explanation:

   $CCl_4\rightleftharpoons C(s)+ 2Cl_2$

Initial  p                     0  0

eq'm   p-x                    0  2x  

Total pressure at equilibrium = p - x + 2x = p + x = 1.9...(1)

$K_p=0.76=\frac{p_{C}\times p^2_{Cl_2}}{p_{CCl_4}}=\frac{0\times(2x)^2}{p-x}$

$K_p=0.76=\frac{4x^2}{p-x}$ ...(2)

Solving equation (1) and (2). we get x = 0.51

p + x = 1.9

p= 1.9 - 0.51 = 1.39 atm

The starting pressure of $CCl_4$ is 1.39 atm.

Answer 2

The starting pressure of ${\text{CC}}{{\text{l}}_{\text{4}}}$ is $\boxed{1.46{\text{ atm}}}$.

Further explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{P(g)}}+{\text{Q(g)}}\rightleftharpoons{\text{R(g)}}+{\text{S(g)}}$

Equilibrium constant in terms of pressure:

It is defined as the ratio of partial pressures of products and that of reactants, both raised to some power that is equal to their respective coefficients in a balanced chemical equation. It is denoted by ${{\text{K}}_{\text{p}}}$

Consider a general balanced reaction,

${\text{aA}}\left(g\right)+{\text{bB}}\left(g\right)\rightleftharpoons{\text{cC}}\left(g\right)+{\text{dD}}\left(g\right)$

The formula to calculate ${{\text{K}}_{\text{p}}}$ is as follows:

${{\text{K}}_{\text{p}}}=\frac{{{{\left[ {{{\text{P}}_{\text{C}}}}\right]}^{\text{c}}}{{\left[{{{\text{P}}_{\text{D}}}}\right]}^{\text{d}}}}}{{{{\left[{{{\text{P}}_{\text{A}}}}\right]}^{\text{a}}}{{\left[{{{\text{P}}_{\text{B}}}}\right]}^{\text{b}}}}}$

Here,

${{\text{P}}_{\text{C}}}$ and ${{\text{P}}_{\text{D}}}$ are the partial pressures of C and D respectively.

${{\text{P}}_{\text{A}}}$ and ${{\text{P}}_{\text{B}}}$ are the partial pressures of A and B respectively.

a and b are the stoichiometric coefficients of A and B respectively.

c and d are the stoichiometric coefficients of C and D respectively.

The decomposition of ${\text{CC}}{{\text{l}}_{\text{4}}}$ occurs as follows:

${\text{CC}}{{\text{l}}_{\text{4}}}\left(g\right)\rightleftharpoons {\text{C}}\left(s\right)+2{\text{C}}{{\text{l}}_{\text{2}}}\left(g\right)$

Consider the change in equilibrium pressure to be x. Therefore, after decomposition, the pressure of ${\text{CC}}{{\text{l}}_{\text{4}}}$ and ${\text{C}}{{\text{l}}_2}$ are p-x and 2p respectively.

The total pressure at equilibrium can be calculated as follows:

$\begin{aligned}{\text{Total pressure}}&=\left( {{\text{p}}-{\text{x}}}\right)+2{\text{x}}\\&={\text{p}}+{\text{x}}\\\end{aligned}$

But total pressure at equilibrium is 1.9 atm. So the above equation becomes,

${\text{p}}+{\text{x}}={\text{1}}{\text{.9}}$                      ...... (1)

Rearrange equation (1) to calculate p.

${\text{p}}={\text{1}}{\text{.9}}-{\text{x}}$               ...... (2)

The formula to calculate the equilibrium constant for the given reaction is as follows:

${{\text{K}}_{\text{p}}}=\frac{{{{\left( {{{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}} \right)}^2}}}{{{{\text{P}}_{{\text{CC}}{{\text{l}}_{\text{4}}}}}}}$                   ...... (3)

Here,

${{\text{K}}_{\text{p}}}$ is the equilibrium constant.

${{\text{P}}_{{\text{CC}}{{\text{l}}_{\text{4}}}}}$ is the pressure of ${\text{CC}}{{\text{l}}_{\text{4}}}$.

${{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}$ is the pressure of ${\text{C}}{{\text{l}}_2}$.

Substitute 0.76 for ${{\text{K}}_{\text{p}}}$, 2p for ${{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}$ and p-x for ${{\text{P}}_{{\text{CC}}{{\text{l}}_{\text{4}}}}}$ in equation (1).

${\text{0}}{\text{.76}} = \frac{{{{\left( {{\text{2x}}} \right)}^2}}}{{\left( {{\text{p}} - {\text{x}}} \right)}}$             ...... (4)

Substitute value of p from equation (2) in equation (4).

$\begin{aligned}{\text{0}}{\text{.76}}&=\frac{{{{\left({{\text{2x}}}\right)}^2}}}{{\left\{{\left({{\text{1}}{\text{.9}}-{\text{x}}}\right)-{\text{x}}}\right\}}}\hfill\\{\text{0}}{\text{.76}}&=\frac{{4{{\text{x}}^2}}}{{1.9-2{\text{x}}}}\hfill\\\end{aligned}$

Simplifying the above equation,

$4{{\text{x}}^2}+1.52{\text{x}}-1.444=0$

Solve for x,

$x=0.44$

Substitute 0.44 for x in equation (2).

$\begin{aligned}{\text{p}}&={\text{1}}{\text{.9}}-{\text{0}}{\text{.44}}\\&={\text{1}}{\text{.46}}\\\end{aligned}$

Therefore the starting pressure of ${\mathbf{CC}}{{\mathbf{l}}_{\mathbf{4}}}$ is 1.46 atm.

Learn more:

1. Sort the solubility of gas will increase or decrease: brainly.com/question/2802008.

2. What is the pressure of the gas?: brainly.com/question/6340739.

Answer details:

Grade: School School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: CCl4, Kp, 1.46 atm, x, p, 0.76, decomposition, C, 2Cl2, equilibrium constant, A, B, C, D, a, b, c, d.