For the reaction $CO (g) + {NH}_{3} (g) ⇌ {HCONH}_{2} (g)$ $"CO"(g) + "NH"_3(g) rightleftharpoons "HCONH"_2(g)$ , if $3.00 atm$ $"3.00 atm"$ of $CO$ $"CO"$ and $1.40 atm$ $"1.40 atm"$ of ${NH}_{3}$ $"NH"_3$ were allowed to react in a closed container, then if $K_{p} = 2.70$ $K_p = 2.70$ at a certain temperature, what is $P_{H C O N H_{2}}$ $P_(HCONH_2)$ at equilibrium?

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For the reaction $CO (g) + {NH}_{3} (g) ⇌ {HCONH}_{2} (g)$ $"CO"(g) + "NH"_3(g) rightleftharpoons "HCONH"_2(g)$ , if $3.00 atm$ $"3.00 atm"$ of $CO$ $"CO"$ and $1.40 atm$ $"1.40 atm"$ of ${NH}_{3}$ $"NH"_3$ were allowed to react in a closed container, then if $K_{p} = 2.70$ $K_p = 2.70$ at a certain temperature, what is $P_{H C O N H_{2}}$ $P_(HCONH_2)$ at equilibrium?

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Answer 1 · 2017-10-29T23:39:21

With everything in a gas phase, we can use the partial pressures as proxies for the molar concentrations.

Explanation:

Set up the ICE Table

$sf(" "" "" "" "" ""CO"" "" "" ""NH"_3" "rightleftharpoons" ""HCONH"_2)$

$sf"Initial"$ $sf(" "" "" "3.00" "" "" "1.40" "" "" "" "0.00)$
$sf"Change"$ $sf(" "" " -x " "" "" "-x" "" "" "" "+x)$
$sf"Equilibrium"$ $sf(" "3 - x " "" "1.4 - x" "" "" "" " x)$

Then put those into the equilibrium expression and solve for x.

$2.70 = x/((3-x)(1.4-x))$

' $x$ ' will be the desired $sf(HCONH_2)$ partial pressure.
You can calculate the equilibrium concentrations of the other gases from it, if desired.

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