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3 years ago

Why are the optimum conditions for the Haber process a temperature of 450°C and pressure of 200 atmospheres?

Chemistry

2 answers:

Novay_Z [31]3 years ago

8 0

Answer:

See explanation.

Explanation:

Hello,

Haber process is defined as the widely acknowledged productive process of ammonia by the reaction:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Which is carried out in gaseous phase. Thus, by means of the Le Chatelier's principle, it is possible to know that its standard enthalpy of reaction is -45.90 kJ/mol (NIST webbook) for which it is an exothermic chemical reaction, for that reason less ammonia will be produced at high temperature, nonetheless, the temperature should not be too low since the reaction rate significantly decrease, therefore, the optimum found temperature is 450 °C.

Moreover, since there are more moles (3+1=4) at the reactants and less moles at the products (2), increasing the pressure of the reaction increases the yield of ammonia, nonetheless, higher pressures involve the purchasing of more expensive equipment to withstand the high-pressures, for that reason, the best found pressure has been set as 200 atm.

Best regards.

Rus_ich [418]3 years ago

6 0

Answer:

Lower temperature and higher pressure favours the forward reaction in the Haber process

Explanation:

The equation of the Haber process for the production of ammonia is;

3H2(g) + N2(g) ⇄2NH3(g)

The Haber process involves the combination of hydrogen gas and nitrogen gas in a ratio of 3:1.

In the Haber process, the forward reaction is exothermic, hence the reverse reaction is endothermic. The implication of this is, as temperature is increased, the position of equilibrium moves to the left, and the yield of ammonia decreases. Decrease in temperature will favour the forward reaction. Hence the process is operated at a lower temperature of 450°C.

Secondly, the total number of volumes on the reactant side is 4 volumes whereas the total number of volumes on the product side is 2 volumes. Increase in pressure will favour the direction producing a lesser total volume. Hence increasing the pressure to about 200 atmospheres will favour the forward reaction thus more ammonia is produced

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Describe how the mass of the product can be calculated when one reactant is in excess

kvasek [131]

On the off chance that one of the reactants is in overabundance yet you don't know which one it is, you have to compute the hypothetical item mass for the both reactants, with a similar item, and whichever has the lower yield is the one you use to precisely depict masses/sums for the condition, since you can't have more than the non-abundance reactant can create.

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3 years ago

A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0°C. What is the new

dolphi86 [110]

Answer: The new pressure of the gas in Pa is 388462

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas at STP = $10^5Pa$

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 700.0 ml

$V_2$ = final volume of gas = 200.0 ml

$T_1$ = initial temperature of gas = 273 K

$T_2$ = final temperature of gas = $30^oC=273+30=303K$

Now put all the given values in the above equation, we get:

$\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}$

$P_2=388462Pa$

The new pressure of the gas in Pa is 388462

6 0

3 years ago

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

3 years ago

Complete combustion of a 17.12mg sample of xylene In oxygen yielded 56.77mg

Veronika [31]

Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=

106.16×1000

17.12

=0.00016moles

Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO

44.01×1000

56.77

=0.0013

Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H

O==

18.02×1000

14.53

=0.0008

Moles ratios

\frac{0.0013}{0.0008}=1.625

0.0008

0.0013

=1.625

\frac{0.0008}{0.0008}=1

0.0008

Hence molecular fomula

The empirical formula is C 4H 5.

The molecular formula C8H10

8 0

2 years ago