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kvv77 [185]
3 years ago
13

A The Haber process is typically carried out at a temperature of approximately 500∘C. What would happen to the rate of the forwa

rd reaction if the temperature were lowered to 100∘C? The reaction rate would . SubmitHintsMy AnswersGive UpReview Part Part B What would happen to the rate of the forward reaction if the concentration of nitrogen were decreased? The reaction rate would . SubmitHintsMy AnswersGive UpReview Part Part C Which of the following would increase the rate of the reverse reaction? Which of the following would increase the rate of the reverse reaction? increasing the concentration of ammonia decreasing the temperature increasing the concentration of nitrogen SubmitHintsMy AnswersGive UpReview Part Part D What will happen to the rates of the forward and reverse reactions when a catalyst is added? What will happen to the rates of the forward and reverse reactions when a catalyst is added? Forward rate increases; reverse rate decreases. Both forward and reverse rates increase. Both forward and reverse rates decrease. Forward rate decreases; reverse rate increases.
Chemistry
1 answer:
Evgen [1.6K]3 years ago
5 0

Answer:

(I) Decrease in temperature from 500oC to 100oC will increase the rate of the forward reaction.

(II) Decreasing the concentration of nitrogen will decrease the reaction rate of the forward reaction.

(III) Increasing the concentration of ammonia will increase the rate of the reverse reaction.

(IV) When a catalyst is added, the rate of the forward reaction increases while that of the reverse reaction decreases.

Explanation:

Chemical equilibrium is a dynamic reversible process where both the forward and reverse reactions occur at the same time and the concentration of all the chemical species in the system remain constant under standard conditions. Equilibrium state is dependent on:

I. Temperature of the reacting system.

II. Concentration of the reacting system.

III. Pressure of the reacting system ( for gases ).

The reaction equation for the Haber process of ammonia production: N2(g) + 3H2(g) <-> 2NH3(g)

According to Le Chatelier's principle that if an external stress or factor is imposed on an equilibrium system, the equilibrium shifts to annul or neutralize the stress:

(I) Decrease in temperature from 500oC to 100oC will increase the rate of the forward reaction because it was originally in the equilibrium state of a higher temperature.

(II) Decreasing the concentration of nitrogen will decrease the reaction rate of the forward reaction by decreasing the formation of ammonia (product).

(III) Increasing the concentration of ammonia will increase the rate of the reverse reaction by preventing its formation from the reactants nitrogen and hydrogen.

(IV) When a catalyst is added, the rate of the forward reaction increases while that of the reverse reaction decreases. This is achieved by the catalyst lowering the activation energy ( the minimum energy required for the reaction to occur ) of the reaction, hence, making it to occur faster.

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A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

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The Chesapeake Bay watershed continues to develop as population in the region grows. Development and urbanization at the cost of natural landscapes can lead to increased pollution of nutrients and sediment to the Bay, especially from stormwater runoff.

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3 years ago
A diploid somatic ("body") cell has 2n = 20 chromosomes. At the end of mitosis, each daughter cell would have ______ chromosomes
kozerog [31]

Answer:

At the end of mitosis, 2n = 20

At the end of meiosis I, n = 10

At the end of meiosis II, n = 10

Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

7 0
3 years ago
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