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Vika [28.1K]
3 years ago
14

Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reactio

n:
2H+(aq) + Cu(s) H2(g) + Cu2+(aq)
Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. You may use the OWL references to find the values you may need in this question.
Chemistry
1 answer:
solmaris [256]3 years ago
7 0

Answer:

3.3 * 10^-12

Explanation:

The balanced equation of the reaction is;

2H+(aq) + Cu(s) ---------> H2(g) + Cu2+(aq)

Hence two electrons were transferred so n=2

E°cell = E°cathode - E°anode

E°cell = 0 V - 0.34 V

E°cell = - 0.34 V

Then;

E°cell = 0.0592/n log K

Substituting values;

- 0.34 = 0.0592/2 log K

- 0.34/0.0296 = log K

-11.486 = log K

K = Antilog (-11.486)

K = 3.3 * 10^-12

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Answer:

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Explanation:

The first step is the <u>calculation of the moles</u> of H_2O and CO_2, so:

114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2

70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O

Now, in 1 mol of CO2 we have 1  mol of C and in 1 mol of H_2O we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C

3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H

Total~grams=~31.25~+~7.82=39.08~g

grams~of~O=60.00~g-~39.08~g=20.92~g~of~O

Now we can <u>convert the grams</u> of O to moles, so:

20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O

The next step is to divide all the mol values by the <u>smallest one</u>:

O=\frac{1.30~mol~O}{1.30~mol~O}=~1

C=\frac{2.6~mol~C}{1.30~mol~O}=~2

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Consider the reaction given below.
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Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

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