In the haber reaction, patented by german chemist fritz haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous

In the haber reaction, patented by german chemist fritz haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. this reaction is now the first step taken to make most of the world's fertilizer. suppose a chemical engineer studying a new catalyst for the haber reaction finds that 671 liters per second of dinitrogen are consumed when the reaction is run at 271c and 0.99atm. calculate the rate at which ammonia is being produced. give your answer in kilograms per second. round your answer to significant digits.

Answers

0.19 kg/s

Explanation:

1) Word equation (given):

•Dinitrogen gas + dihydrogen gas → gaseous ammonia.

2) Chemical equation:

•N₂ (g) + 3H₂ (g) → 2NH₃ (g)

3) Mole ratios:

•1 mole N₂ (g) : 3 mole H₂ (g) : 2 mole NH₃ (g)

4) Reaction rates:

•Rate of dinitrogen consumption: r₁ = n₁ / t (moles/s)

•Rate of ammonia production: r₂ = n₂ / t (moles/s)

•Due to the stoichiometric ratios: r₂ = 2 × r₁

5) Calculate r₁:

•284 liter / s

•PV = nRT ⇒ n = PV / (RT)

•Divide by time, t: n/t = P (V/t) / (RT)

•Substitute V/t = 284 liter/s, P = 0.75 atm, and T = 196 +273.15K = 469.15K

r₁ = n₁ / t = (0.75 atm) (284 liter/s) / [ (0.08206 atm-liter/K-mol) (469.15k) ]

= 5.53 moles/s

6) Calcualte r₂

r₂ = 2 × r₁ = 2 × 5.53 mole/s = 11.06 mole/s

7) Convert rate in mole/s to rate in kg/s

mass in grams = molar mass × number of molesmolar mass of NH₃ = 17.03 g/molmass = 17.03 g/mol × 11.06 /s = 188.4 g/sConvert 188.4 g/s to kg/s: 0.1884 kg/sCorrect number of significan digits: 2 (since the pressure is given with 2 significant figures. 0.19 kg/s

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

frac{P_1V_1}{T_1} =frac{P_2V_2}{T_2}

frac{.35times 1572}{495} =frac{1times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

The question is incomplete, complete question is :

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 348 liters per second of dinitrogen are consumed when the reaction is run at 205°C and 0.72 atm. Calculate the rate at which ammonia is being produced.

The rate of production of ammonia is 217.08 grams per second.

Explanation:

N_2+3H_2rightarrow 2NH_3

Volume of dinitrogen used in a second = 348 L

Temperature of the gas = T = 205°C = 205+273 K = 478 K

Pressure of the gas = P = 0.72 atm

Moles of dinitrogen = n

n=frac{PV}{RT}=frac{0.72 atmtimes 348 L}{0.0821 atm L/mol Ktimes 478 K}=6.385 mol

According to reaction, 1 mole of dinitriogen gives 2 mole of ammonia.Then 6.385 moles of dinitrogen will give:

frac{2}{1}times 6.385 mol=12.769 mol

Mass of 12.769 moles of ammonia;

12.769 mol 17 g/mol = 217.08 g

217.08 grams of ammonia is produced per second.So, the rate of production of ammonia is 217.08 grams per second.

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

N_2+3H_2rightarrow 2NH_3 Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

PV=nRT

n=frac{PV}{RT}=frac{0.88 atmtimes 505 L}{0.0821 atm l/mol Ktimes 445.15 K}=12.1597 mol

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

frac{2}{1}times 12.1597 mol=24.3194 mol of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

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