Question

Ammonia burns in air to form nitrogen dioxide and water. $4{NH}_3(g) + 7{O}_2(g) \longrightarrow 4{NO}_2(g) + 6{H}_2{O}(l) $ If 8 moles of $NH_3$ are reacted with 14 moles of $O_2$ in a rigid container with an initial pressure of 11 atm, what is the partial pressure of $NO_2$ in the container when the reaction runs to completion? (Assume constant temperature)

• A. 4 atm
• B. 6 atm
• C. 11 atm
• D. 12 atm

Hint:

To calculate the partial pressure of a gas in a rigid container, use the mole fraction of the gas and multiply it by the total pressure.

Answer 1

The Correct Option is A

First, we use the balanced chemical equation: \[ 4{NH}_3(g) + 7{O}_2(g) \longrightarrow 4{NO}_2(g) + 6{H}_2{O}(l) \] This shows that for every 4 moles of NH3, 4 moles of NO2 are produced. Since we start with 8 moles of NH3, we will produce 8 moles of NO2. The total initial moles of gas in the container are \( 8 { moles of NH}_3 + 14 { moles of O}_2 = 22 { moles of gas} \). The initial pressure is 11 atm, so the total pressure is proportional to the total moles of gas. The final pressure is proportional to the moles of NO2 formed, which is 8 moles. Thus, the partial pressure of NO2 is: \[ P_{{NO}_2} = \frac{8}{22} \times 11 = 4 { atm}. \] Therefore, the correct answer is (a) 4 atm.