# For the reaction "CO"(g) + "NH"_3(g) rightleftharpoons "HCONH"_2(g), if "3.00 atm" of "CO" and "1.40 atm" of "NH"_3 were allowed to react in a closed container, then if K_p = 2.70 at a certain temperature, what is P_(HCONH_2) at equilibrium?

Oct 29, 2017

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

#### Explanation:

Set up the ICE Table

$\textsf{{\text{ "" "" "" "" ""CO"" "" "" ""NH"_3" "rightleftharpoons" ""HCONH}}_{2}}$

$\textsf{\text{Initial}}$$\textsf{\text{ "" "" "3.00" "" "" "1.40" "" "" "" } 0.00}$
$\textsf{\text{Change}}$$\textsf{\text{ "" " -x " "" "" "-x" "" "" "" } + x}$
$\textsf{\text{Equilibrium}}$$\textsf{\text{ "3 - x " "" "1.4 - x" "" "" "" } x}$

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

$2.70 = \frac{x}{\left(3 - x\right) \left(1.4 - x\right)}$

'$x$' will be the desired $\textsf{H C O N {H}_{2}}$ partial pressure.
You can calculate the equilibrium concentrations of the other gases from it, if desired.