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Alex787 [66]
4 years ago
3

How does temperature affect the copper (II) chloride equilibrium? Is the forward reaction (color changing from blue to green) en

dothermic or exothermic? Justify your choice with experimental evidence i.e color changes in the video for Part B.
Chemistry
1 answer:
hichkok12 [17]4 years ago
3 0

Answer:

See explanation

Explanation:

A popular experiment that describes the effect of heat on the position of equilibrum is the change of colour when copper II chloride is heated.

As the solution is heated, it's colour changes from blue to green, this implies the the colour change (blue to green) is an endothermic process (equilibrum position shifts to the right with increase in temperature)

The equilibrum is represented by the equation;

[Cu(H2O)6]^2+(aq) + 4Cl^-(aq)<------>[CuCl4]^2-(aq) + 6H2O(l) ∆H=positive

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Answer:

94.58 g of H_2O

Explanation:

For this question we have to start with the reaction:

H_2~+~O_2~->~H_2O

Now, we can balance the reaction:

2H_2~+~O_2~->~2H_2O

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1) Find the moles of each reactive, using the molar mass of each compound (H_2~=~2~g/mol~~O_2=~32~g/mol ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for H_2 and "1" for O_2).

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32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2

84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2

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<u />

\frac{15.87~mol~H_2}{2}=7.94

\frac{2.62~mol~of~O_2}{1}=2.62

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32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O

We have to remember that the molar ratio between H_2O and H_2 is 2:2 and the molar mass of H_2O is 18 g/mol.

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