Question #19cd0

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Question #19cd0

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Answer 1 · 2015-11-19T20:53:53

$2.42 \cdot 10^{- 1} M$ $2.42 * 10^(-1)"M"$

Explanation:

Start with the balanced chemical equation for this equilibrium reaction

$2 {CH}_{2} {Cl}_{2(g]} ⇌ {CH}_{4(g]} + {CCl}_{4(g]}$ $color(red)(2)"CH"_2"Cl"_text(2(g]) rightleftharpoons "CH"_text(4(g]) + "CCl"_text(4(g])$

By definition, the equilibrium constant, $K_{c}$ $K_c$ , is equal to the ratio between the product of the equilibrium concentrations of the products and the equilibrium concentration of the reactant, all raised to the power of their respective stoichiometric coefficients.

Mathematically, $K_{c}$ $K_c$ will be equal to

$K_{c} = \frac{[{CH}_{4}] \cdot [{CCl}_{4}]}{{[{CH}_{2} {Cl}_{2}]}_{2}^{2}}$ $K_c = ( ["CH"_4] * ["CCl"_4])/(["CH"_2"Cl"_2]^color(red)(2))$

Keep in mind, these are all equilibrium concentrations used in the expression for $K_{c}$ $K_c$ .

Now, you know that this reaction has a $K_{c}$ $K_c$ equal to $10.5$ $10.5$ at a temperature of $350 K$ $"350 K"$ .

Moreover, you know that at equilibrium, the reaction vessel contains

$[{CH}_{2} {Cl}_{2}] = 2.36 \cdot 10^{- 2} M$ $["CH"_2"Cl"_2] = 2.36 * 10^(-2)"M"$

and

$[{CH}_{4}] = 2.42 \cdot 10^{- 2} M$ $["CH"_4] = 2.42 * 10^(-2)"M"$

Even without doing any calculations, you could predict that the concentration of carbon tetrachloride, ${CCl}_{4}$ $"CCl"_4$ , will be about $10$ $10$ times bigger than that of the other product, methane.

That happens because the value of $K_{c}$ $K_c$ , which is greater than one, tells you that, at this temperature at least, the reaction favors the products.

Since the equilibrium concentrations of the reactant and of one of the products are approximately equal, the magnitude of $K_{c}$ $K_c$ must have an impact on the concentration of methane.

You can determine the equilibrium concentration for ${CCl}_{4}$ $"CCl"_4$ by rearranging the equation for $K_{c}$ $K_c$

$K_{c} = \frac{[{CH}_{4}] \cdot [{CCl}_{4}]}{{[{CH}_{2} {Cl}_{2}]}_{2}^{2}} \Rightarrow [{CCl}_{4}] = K_{c} \cdot \frac{{[{CH}_{2} {Cl}_{2}]}_{2}^{2}}{[{CH}_{4}]}$ $K_c = ( ["CH"_4] * ["CCl"_4])/(["CH"_2"Cl"_2]^color(red)(2)) implies ["CCl"_4] = K_c * (["CH"_2"Cl"_2]^color(red)(2))/(["CH"_4])$

Plug in your values to get

$["CCl"_4] = 10.5 * ( (2.36 * 10^(-2))^color(red)(2)"M"^color(red)(cancel(color(black)(2))))/( 2.42 * 10^(-2)color(red)(cancel(color(black)("M")))) = 24.166 * 10^(-2)"M"$

Rounded to three sig figs, the answer will be

$["CCl"_4] = color(green)(2.42 * 10^(-1)"M")$

Indeed, the initial predict appears to be correct, you do have about $10$ time more carbon tetrachloride than methane.

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Question #19cd0

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