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Sladkaya [172]
3 years ago
15

In an aqueous chloride solution cobalt(II) exists in equilibrium with the complex ion CoCl42-. Co2 (aq) is pink and CoCl42-(aq)

is blue. At Low Temperature the pink color pre-dominates. At High Temperature the blue color is strong. If we represent the equilibrium as:
CoCl4^2-(aq) <--> Co2+(aq) + 4Cl-(aq)
We can conclude that:___.
1. This reaction is:___.
A. Exothermic
B. Endothermic
C. Neutral
2. When the temperature is decreased the equilibrium constant, K:____.
A. Increases
B. Decreases
C. Remains the same.
3. When the temperature is decreased the equilibrium concentration of Co2:______.
A. Increases
B. Decreases
C. Remains the same.
Chemistry
1 answer:
kolezko [41]3 years ago
4 0

Answer:

1. This reaction is <u>(A) Exothermic .</u>

2. When the temperature is decreased the equilibrium constant, K: <u>(A) Increases</u>

3.When the temperature is decreased the equilibrium concentration of Co2:<u> (A) Increases</u>

Explanation:

CoCl4^2-(aq)  Co_2+(aq) + 4Cl^-(aq)

1. The pink color predominates at low temperatures, indicating that the commodity is preferred.

This is a reaction that is <u>exothermic.</u>

2. As the decrease in the temperature , the equilibrium constant , K ;

      equilibrium constant = \frac{product}{reactant} =\frac{[CO^2^+][Cl^-^4]}{CoCl^2^-_4}

As the temperature drops, the concentration ofCl^- and CO^2^+ rises, and K rises as well , thus it <u>increases </u>.

3. The equilibrium concentration of CO^2^+ decreases as the temperature decreases:

When the temperature is lowered, the equilibrium shifts to the right , that is it <u>increases.</u>

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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
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Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

and,

The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

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