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monitta
3 years ago
15

Nitrogen gas (N₂) reacts with hydrogen gas (H₂) to form ammonia (NH₃). At 200°C in a closed container, 1.0 atm of nitrogen gas i

s mixed with 2.0 atm of hydrogen gas. At equilibrium, the total pressure is 1.9 atm. Calculate the partial pressure of hydrogen gas at equilibrium.
Chemistry
1 answer:
Pie3 years ago
8 0

Answer:

The partial pressure of hydrogen gas at equilibrium is 1.26 atm

Explanation:

Let's use the molar fraction to solve this:

Molar fraction = Moles of gas / Total moles

Molar fraction = Gas pressure / Total pressure

Without equilibrium, we can think that the total system pressure is the sum of the partial pressures of each gas.

1 atm N₂ + 2 atm H₂ = 3 atm

Molar fraction for H₂ = 2 atm / 3atm → 0.66

Let's replace the molar fraction in equilibrium

Gas pressure / 1.9 atm = 0.66

Gas pressure = 1.26atm

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sleet_krkn [62]

Answer:

p_{H_2O}=2.00atm

Explanation:

Hello!

In this case, according to the following chemical reaction:

2H_2+O_2\rightarrow 2H_2O

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm

Best regards!

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