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Fittoniya [83]
3 years ago
14

Decreasing the temperature of the reaction 3H2 + N2<----->2NH3. In this reaction, the product absorbs heat. WHICH WAY WILL

THE REACTION SHIFT ???
Chemistry
1 answer:
AURORKA [14]3 years ago
7 0

3H_{2}+N_{2}⇔2NH_{3}

Decreasing the temperature of the reaction,the reaction shifts forward.

The explanation is given below.

Explanation:

If the temperature of the reaction mixture is increased,then the equilibrium will shift to decrease the temperature.

If the temperature of the reaction mixture is decreased,then the equilibrium will shift to increase the temperature.

During the formation of the ammonia,it gives off heat.So it is an exothermic reaction.

3H_{2}+N_{2}⇔2NH_{3}

A decrease in the temperature favors the reaction that is exothermic (the forward reaction)because it produces energy.Therefore,if the temperature is decreased,the yield of the ammonia increases.

<em>Therefore if the temperature is increased,the reaction shifts forward and the yield of the ammonia increases and it is an exothermic reaction.</em>

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Suppose a system returns to its original overall, internal energy ((∆U) after the following changes. In step 1, 25 J of work is
Brums [2.3K]

Answer:

Heat transfer in step 2 = 47.75 J

Explanation:

Internal energy = heat + work done

U = Q + W

In a cyclic process the total internal energy change of the system = 0.

In the process there are two steps. The total heat exchange in the process is the sum of heat exchanges in the two processes.

We have to find the heat exchange in step 2.

In step 1,

W = 1.25 J          Q = -37 J

U_{1} = -37 + 1.25 = -35.75 J

In step 2, the internal energy change will be negative of that in step 1.

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W = -12 J

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35.75 = Q -12

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6 0
4 years ago
How do solve for #13?What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?
exis [7]

What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

The change in Boiling Point of water can be calculated using this formula:

ΔTb = i * Kb * m

Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.

The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m

The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1

The molal concentration of the solution can be found using this formula:

molality = moles of sucrose/kg of water

molality = 1.000 mol / 1.000 kg of water

molality = 1 m

Now that we know all the values, we can use the formula to find the change in the boiling point of water:

ΔTb = i * Kb * m

ΔTb = 1 * 0.515 °C/m * 1 m

ΔTb = 0.515 °C

Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:

Boiling point solution = Boiling point of water + ΔTb

Boiling point solution = 100.000 °C + 0.515 °C

Boiling point solution = 100.515 °C

Answer: The boiling point of the solution is 100.515 °C.

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