Step 1: Identify the limiting reactant
We can use partial pressures as a replacement for moles because $P ∝ n$.
$color(white)(mmmmmm)"N"_2color(white)(m) + color(white)(m)"3H"_2 → "2NH"_3$
$P"(atm):"color(white)(mll) 2.73color(white)(mmm)2.73$
$"Divide by:" color(white)(mll)1color(white)(mmmml)3$
$"Moles rxn:" color(white)(ll)2.73color(white)(mmm)0.910$
Hydrogen is the limiting reactant because it gives the fewest moles of reaction,
Step 2: Calculate the partial pressures after the reaction
$color(white)(mmmmmm)"N"_2color(white)(m) + color(white)(m)"3H"_2 → "2NH"_3$
$"I:/atm:"color(white)(mlm) 2.73color(white)(mmm)2.73color(white)(mmll)0$
$"C:/atm:" color(white)(ml)"-0.910"color(white)(mml)"-2.73"color(white)(ml)"+1.82"$
$"E:/atm:" color(white)(mm)1.82color(white)(mmmm)0color(white)(mmll)1.82$
The product mixture will contain ammonia and unreacted nitrogen, each with a partial pressure of 1.82 atm.
$P_"total" = P_"N₂" + P_"NH₃" = "1.82 atm + 1.82 atm" = "3.64 atm"$
$P_"NH₃" = "1.82 atm"$.
Step 4. Now let the cylinder contract and calculate the new volume.
Remember that we are using partial pressures as a substitute for moles.
We use :
$color(blue)(bar(ul(|color(white)(a/a) n_1/V_1 = n_2/V_2 color(white)(a/a)|)))" "$
$n_1 = "5.46 atm"; V_1 = "15.0 L"$
$n_2 = "3.64 atm"; V_2 = ?"$
$V_2 = V_1 × n_2/n_1 = "15.0 L" × (3.64 color(red)(cancel(color(black)("atm"))))/(5.46 color(red)(cancel(color(black)("atm")))) = "10.0 L"$
