# How do you calculate the partial pressure of ammonia in the container after the reaction has reached completion? How do you calculate the volume of the container after the reaction has reached completion?

## Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia gas (NH3). You have nitrogen and hydrogen gases in a 15.0 L container fitted with a movable piston (the piston allows the container volume to change so as to keep the pressure constant inside the container). Initially the partial pressure of each reactant gas is 2.73 atm. Assume the temperature is constant and the reaction goes to completion.

Oct 27, 2016

The partial pressure of ammonia is 1.82 atm. The new volume is 10.0 L.

#### Explanation:

Step 1: Identify the limiting reactant

We can use partial pressures as a replacement for moles because P ∝ n.

$\textcolor{w h i t e}{m m m m m m} {\text{N"_2color(white)(m) + color(white)(m)"3H"_2 → "2NH}}_{3}$
$P \text{(atm):} \textcolor{w h i t e}{m l l} 2.73 \textcolor{w h i t e}{m m m} 2.73$
$\text{Divide by:} \textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{m m m m l} 3$
$\text{Moles rxn:} \textcolor{w h i t e}{l l} 2.73 \textcolor{w h i t e}{m m m} 0.910$

Hydrogen is the limiting reactant because it gives the fewest moles of reaction,

Step 2: Calculate the partial pressures after the reaction

$\textcolor{w h i t e}{m m m m m m} {\text{N"_2color(white)(m) + color(white)(m)"3H"_2 → "2NH}}_{3}$
$\text{I:/atm:} \textcolor{w h i t e}{m l m} 2.73 \textcolor{w h i t e}{m m m} 2.73 \textcolor{w h i t e}{m m l l} 0$
$\text{C:/atm:" color(white)(ml)"-0.910"color(white)(mml)"-2.73"color(white)(ml)"+1.82}$
$\text{E:/atm:} \textcolor{w h i t e}{m m} 1.82 \textcolor{w h i t e}{m m m m} 0 \textcolor{w h i t e}{m m l l} 1.82$

The product mixture will contain ammonia and unreacted nitrogen, each with a partial pressure of 1.82 atm.

${P}_{\text{total" = P_"N₂" + P_"NH₃" = "1.82 atm + 1.82 atm" = "3.64 atm}}$

${P}_{\text{NH₃" = "1.82 atm}}$.

Step 4. Now let the cylinder contract and calculate the new volume.

Remember that we are using partial pressures as a substitute for moles.

color(blue)(bar(ul(|color(white)(a/a) n_1/V_1 = n_2/V_2 color(white)(a/a)|)))" "
${n}_{1} = \text{5.46 atm"; V_1 = "15.0 L}$
${n}_{2} = \text{3.64 atm"; V_2 = ?}$
V_2 = V_1 × n_2/n_1 = "15.0 L" × (3.64 color(red)(cancel(color(black)("atm"))))/(5.46 color(red)(cancel(color(black)("atm")))) = "10.0 L"