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ICE Princess25 [194]
2 years ago
12

Question 1 (Matching Worth 3 points)

Chemistry
1 answer:
Misha Larkins [42]2 years ago
5 0

Answer:

Question 1: 1) Increasing the pressure          C) Shift to the right

                   2) Removing hydrogen gas        A) Shift to the left  

                   3) Adding a catalyst                     B) No effect

Question 2: This reaction is exothermic because the system shifted to the right on cooling.

Question 3: Shift it toward the reactants.

Question 4: Adding more of gas C to the system.

Question 5: It will shift toward the reactant side because the reactant side has one more mole of gas than the product side.

Question 6: True.

Question 7: there is no suitable choice is provided.

We can shift the equilibrium toward the right via:

Increasing N2O3 concentration,

decreasing NO and/or NO2 concentration,

decreasing the pressure,

lowering the T (cooling the system).

Explanation:

<em>Question 1: Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). </em>

<em>Match Term Definition </em>

<em>Removing ammonia A) No effect </em>

<em>Removing hydrogen gas B) Shift to the left </em>

<em>Adding a catalyst C) Shift to the right</em>

  • <em>Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>

<u><em>1) Increasing the pressure:</em></u>

  • When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
  • The reactants side (left) has 4.0 moles of gases and the products side (right) has 2.0 moles of gases.
  • So, increasing the pressure will shift the reaction to the side with lower moles of gas (right side).

<u><em>so, the right match is: C) Shift to the right.</em></u>

<u><em>2) Removing hydrogen gas:</em></u>

  • Removing hydrogen gas will decrease the concentration of the products side, so the reaction will be shifted to the lift side to suppress the decrease in the concentration of hydrogen gas by removing.

<u><em>so, the right match is: A) Shift to the left.</em></u>

<u><em>3) Adding a catalyst:</em></u>

  • Catalyst increases the rate of the reaction without affecting the equilibrium position.
  • Catalyst increases the rate via lowering the activation energy of the reaction.
  • This can occur via passing the reaction in alternative pathway (changing the mechanism).
  • The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
  • in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
  • with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.

<em><u>so, the right match is: B) No effect.</u></em>

<em><u></u></em>

Question 2: Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.

2NO2(g) ⇌ N2O4(g)

When the light brown color equilibrium mixture was moved from room temperature to a lower temperature, the mixture turned lighter brown in color. Which of the following conclusions about this equilibrium mixture is true?

  • Moving from room temperature to a lower temperature, means that the T is decreased.
  • The mixture turned lighter brown in color, means that the reaction is shifted towards the products side.
  • Moving towards the product side, means that we decreased the concentration of the product side, which ,means that the heat is evolved from the reaction and the reaction is exothermic.

<u><em>So, the right choice is: This reaction is exothermic because the system shifted to the right on cooling.</em></u>

<u><em></em></u>

<em>Question 3: According to Le Châtelier's principle, how will a decrease in concentration of a reactant affect the equilibrium system?</em>

<em><u> </u></em>

  • A decrease in the reactants concentration will disturb the equilibrium position of the system. To attain the equilibrium again, the system will be shifted to the lift (reactants side) to increases the concentration of the reactants and attain the equilibrium again.

<u><em>So, the right choice is: Shift it toward the reactants.</em></u>

<u><em /></u>

<u><em>Note: The answer of Q 4, 5, 6 & 7 and all answers are in the attached word file.</em></u>

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Then, we multiply this number of moles by the molar weight of cfa which is equal to 88 g/mol. 
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Answer;

= 18.24

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The ratio of N and O in the formula NO2 IS 1:2

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= 0.03125 × 2 = 0.0625 moles

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0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
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Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

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We are given with
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Answer:

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