A vessel at 1000K contains #"CO"_2# with a pressure of 0.5atm. Some of the #"CO"_2# is converted into #"CO"# on the addition of graphite. If the total pressure at equilibrium is 0.8atm, the value of #K_p# is?
a) 1.8 atm B)3 atm c) 0.3 atm d) 0.18 atm
a) 1.8 atm B)3 atm c) 0.3 atm d) 0.18 atm
1 Answer
Explanation:
Start by writing the balanced chemical equation that describes this equilibrium reaction
#"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
At equilibrium, the partial pressure of carbon dioxide will be
#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
This means that if
Consequently, you can say that at equilibrium, the partial pressure of carbon monoxide will be
#P_("CO") = (color(blue)(2) * x)" atm"#
Now, the total pressure at equilibrium is said to be equal to
#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 = ("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.