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k0ka [10]
3 years ago
13

Answer the questions for the following reaction at chemical equilibrium: Cl2(g) + KBr(aq) <----> KCl(aq) + Br2(l) + heat 1

. Which reaction (forward, backward, or neither) would be favored by increasing the temperature? 2. Which reaction (forward, backward, or neither) would be favored by adding CaO? 3. Which reaction (forward, backward, or neither) would be favored if the amount of chlorine gas (Cl2 gas) was decreased? 4. Which reaction (forward, backward, or neither) would be favored if the pressure was increased? 5. Which reaction (forward, backward, or neither) would be favored if more Cl2 gas is added?
Chemistry
2 answers:
ollegr [7]3 years ago
8 0

Answer:

1. Backward.

2. Neither.

3. Backward.

4. Forward.

5. Forward.

Explanation:

Hello,

This exercise is solved by applying the Le Chatelier's principle for each statement as follows:

1. At first, by increasing the temperature we are kind of adding heat which is a product in this chemical reaction, therefore the backward reaction is favored.

2. Next, as there are no species regarding to the CaO, neither reaction is favored by its addition.

3. Afterwards, the withdrawal of chlorine gas favors the backward reaction as the reactants concentration is lowered.

4. Then, since the change in the moles for this chemical reaction is given by:

\Delta \nu=-1

Whereas \Delta \nu accounts for the change in the stoichiometric coefficient for each gaseous species (negative for reactants and positive for products). In such a way, as the pressure is increased the equilibrium will shift to the side with less gaseous molecules based on \Delta \nu, that is forward.

5. Finally, as chlorine is being added, the equilibrium will shift forward to formation of more products.

Best regards.

const2013 [10]3 years ago
4 0

Answer:

1.Backward

2.Neither

3.Backward

4.Forward

5.Forward

Explanation:

When we increase the temperature

Then, by Lechatelier principle

It states that if a constraints such as concentration of reactants,change in volume , pressure,temperature  is applied to a system of equilibrium ,the equilibrium shift in a direction in which oppose the effect of constraints.

Using Lechatelier principle

1.The equilibrium shift in backward direction when we increase the temperature.

2.When we add CaO then it is not reactant of given reaction or not a product therefore, when we add CaO then it does not effect euilibrium condition.

Then , the equilibrium neither shift in any direction.

3.When we decrease the concentration of chlorine gas then the number of

Then the equilibrium shift in backward direction by Le chetelier principle.

Hence , the reaction would be favored backward  if the amount of chlorine gas was decreased.

4.When we increase the pressure then the equilibrium shift in forward direction.

Hence, the reaction would be favored in forward reaction when we increase the pressure.

5.When we increase concentration of any  reactant then the equilibrium shift in forward direction.

Hence, when we add more chlorine gas then the reaction would be favored forward reaction.

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\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

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We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

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Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

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We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

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