A vessel at 1000K contains ${CO}_{2}$ $"CO"_2$ with a pressure of 0.5atm. Some of the ${CO}_{2}$ $"CO"_2$ is converted into $CO$ $"CO"$ on the addition of graphite. If the total pressure at equilibrium is 0.8atm, the value of $K_{p}$ $K_p$ is?

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A vessel at 1000K contains ${CO}_{2}$ $"CO"_2$ with a pressure of 0.5atm. Some of the ${CO}_{2}$ $"CO"_2$ is converted into $CO$ $"CO"$ on the addition of graphite. If the total pressure at equilibrium is 0.8atm, the value of $K_{p}$ $K_p$ is?

a) 1.8 atm B)3 atm c) 0.3 atm d) 0.18 atm

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Answer 1 · 2016-11-08T00:55:58

$K_{p} = 1.8 atm$ $K_p = "1.8 atm"$

Explanation:

Start by writing the balanced chemical equation that describes this equilibrium reaction

${CO}_{2 (g)} + C_{(s)} \to 2 {CO}_{(g)}$ $"CO"_ (2(g)) + "C"_ ((s)) -> color(blue)(2)"CO"_ ((g))$

Now, this reaction takes place at constant volume and temperature, which means that you can use a decrease in partial pressure to be proportional to a decrease in moles.

Now, you know that some of the carbon dioxide is converted to carbon monoxide.

This means that the number of moles of carbon dioxide present in the vessel decreases by a value, let's say $x$ $x$ . You can thus say that the partial pressure of carbon dioxide will decrease by $x$ $x$ .

At equilibrium, the partial pressure of carbon dioxide will be

$P_{{CO}_{2}} = (0.5 - x) atm$ $P_("CO"_2) = (0.5 - x)" atm"$

Now, you know that carbon monoxide was not present in the vessel before the reaction took place, which, of course, implies that its initial partial pressure is zero.

Notice that the reaction produces $2$ $color(blue)(2)$ molecules of carbon monoxide for every $1$ $1$ molecule of carbon dioxide that takes part in the reaction.

This means that if $x$ $x$ moles of carbon dioxide react, the reaction will produce twice as many moles of carbon monoxide.

Consequently, you can say that at equilibrium, the partial pressure of carbon monoxide will be

$P_{CO} = (2 \cdot x) atm$ $P_("CO") = (color(blue)(2) * x)" atm"$

Now, the total pressure at equilibrium is said to be equal to $0.8 atm$ $"0.8 atm"$ . This means that the partial pressures of the two gases must add up to give $0.8 atm$ $"0.8 atm"$ $\to$ $->$ think Dalton's Law of Partial Pressures here.

$P_{total} = P_{{CO}_{2}} + P_{CO}$ $P_ "total" = P_("CO"_ 2) + P_"CO"$

You will thus have

$0.8 color(red)(cancel(color(black)("atm"))) = (0.5 - x) color(red)(cancel(color(black)("atm"))) + (color(blue)(2)x)color(red)(cancel(color(black)("atm")))$

$0.8 = 0.5 + x implies x= 0.3$

The equilibrium partial pressures of the two gases will be

$P_("CO"_2) = 0.5 - 0.3 = "0.2 atm"$

$P_("CO") = color(blue)(2) * 0.3 = "0.6 atm"$

By definition, the equilibrium constant for this reaction, $K_p$ , will be -- keep in mind that solids are not added to the expression of the equilibrium constant!

$K_p = ("CO")^color(blue)(2)/(("CO"_2))$

Plug in your values to find

$K_p = ("0.6 atm")^color(blue)(2)/("0.2 atm") = color(green)(bar(ul(|color(white)(a/a)color(black)("1.8 atm")color(white)(a/a)|)))$

I'll leave the answer rounded to two sig figs.

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A vessel at 1000K contains ${CO}_{2}$ $"CO"_2$ with a pressure of 0.5atm. Some of the ${CO}_{2}$ $"CO"_2$ is converted into $CO$ $"CO"$ on the addition of graphite. If the total pressure at equilibrium is 0.8atm, the value of $K_{p}$ $K_p$ is?

a) 1.8 atm B)3 atm c) 0.3 atm d) 0.18 atm

1 Answer

Explanation:

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A vessel at 1000K contains "CO"_2CO2"CO"_2 with a pressure of 0.5atm. Some of the "CO"_2CO2"CO"_2 is converted into "CO"CO"CO" on the addition of graphite. If the total pressure at equilibrium is 0.8atm, the value of K_pKpK_p is?

a) 1.8 atm B)3 atm c) 0.3 atm d) 0.18 atm

1 Answer

Explanation:

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A vessel at 1000K contains ${CO}_{2}$ $"CO"_2$ with a pressure of 0.5atm. Some of the ${CO}_{2}$ $"CO"_2$ is converted into $CO$ $"CO"$ on the addition of graphite. If the total pressure at equilibrium is 0.8atm, the value of $K_{p}$ $K_p$ is?