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12345 [234]
2 years ago
6

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Chemistry
1 answer:
shusha [124]2 years ago
3 0

This problem is asking for the rate of disappearance of gaseous nitrogen, given the rate of appearance of ammonia and the chemical reaction. At the end, the result turns out to be -0.228 M/s.

<h3>Rates of appearance and disappearance</h3>

In chemical kinetics, one of the most relevant calculations are based on rates of appearance and disappearance of chemical species in a chemical reaction. This can be calculated via rate portions based on the stoichiometric coefficients in the reaction.

Thus, for this problem, one can write:

\frac{r_{N_2}}{-1} =\frac{r_{H_2}}{-3} =\frac{r_{NH_3}}{2}

Where the rate of appearance or disappearance is divided by the stoichiometric coefficient. Therefore, one can solve for the rate of disappearance of N2 with:

\frac{r_{N_2}}{-1} =\frac{r_{NH_3}}{2}\\\\r_{N_2}=\frac{r_{NH_3}*-1}{2}\\\\r_{N_2}=\frac{0.456M/s*-1}{2}\\\\r_{N_2}=-0.228M/s

Learn more about chemical kinetics: brainly.com/question/26351746

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3 years ago
Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H
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Answer:

36.37% is the percent yield of the reaction.

Explanation:

4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)

1)0.650 L nitrogen gas  , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P=  1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T =  295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol

2) Moles of ammonia gas=\frac{2.53 g}{17 g/mol}=0.1488 mol

Moles of oxygen gas =\frac{3.53 g}{32 g/mol}=0.1101 mol

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

\frac{4}{3}\times 0.1101 mol=0.1468 mol of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

\frac{2}{3}\times 0.1101 mol=0.0734 mol of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100

Percentage yield of the reaction:

\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%

36.37% is the percent yield of the reaction.

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