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Gwar [14]
3 years ago
10

Consider this reaction: 2Al(s) + 3 CuCl2(aq) → 2AlCl3(aq) + 3 Cu(s) If the concentration of CuCl2 drops from 1.000 M to 0.655 M

in the first 30.0 s of the reaction, what is the average rate of reaction over this time interval?
Chemistry
2 answers:
Yuliya22 [10]3 years ago
6 0

Answer:

r=-0.0115M/s

Explanation:

Hello,

In this case, the rate is computed as:

r=\frac{C_F-C_0}{T_F-T_0}

As the reaction is analyzed during the first 30 seconds, the average rate will be:

r=\frac{0.655M-1.000M}{30s-0s}=-0.0115M/s

Which negative since the cupric chloride is a reagent.

Best regards.

blondinia [14]3 years ago
4 0

Answer:

rate = 0.00383 M/s

Explanation:

The rate of reaction refers to the speed at which the products are formed from the reactants in a chemical reaction

2Al(s) + 3 CuCl2(aq) → 2AlCl3(aq) + 3 Cu(s)

The equation for reaction rate in this chemical reaction (for CuCl₂)  is the following one (The value 1/3 comes from the coefficient in the chemical equation for the reaction):

\text{rate of reaction}  = - \frac{1}{\text{coefficient}} \frac{Reactant}{Time}

Here, the negative sign is used to indicate the decreasing concentration of the reactant.

rate= -\frac{1}{3} \times \frac{C2-C1}{t2-t1}

rate = - 1/3 [0.655 - 1.0 / 30]

rate = -1/3[-0.0115]

rate = 3.83 x 10-3 M/s

rate = 0.00383 M/s

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Read 2 more answers
The reaction 2CH4(g)⇌C2H2(g)+3H2(g) has an equilibrium constant of K = 0.154. If 6.30 mol of CH4, 4.20 mol of C2H2, and 11.15 mo
boyakko [2]

Answer:

C₂H₂ + 3H₂ ⟶ 2CH₄  

Explanation:

The initial concentrations are:

[CH₄] = 6.30 ÷ 6.00 =   1.05  mol·L⁻¹

[C₂H₂] = 4.20 ÷ 6.00 = 0.700 mol·L⁻¹

   [H₂] = 11.15 ÷  6.00 =  1.858 mol·L⁻¹

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I/mol·L⁻¹:    1.05     0.700   1.858

Q = \dfrac{\text{[C$_{2}$H$_{2}$][H$_{2}$]}^{3}}{\text{[CH$_{4}$]}^{2}} = \dfrac{ 0.700\times 1.858^{3}}{1.05^{2}}= 4.07

Q > K

That means we have too many products.

The reaction will go to the left to get rid of the excess products.

C₂H₂ + 3H₂ ⟶ 2CH₄

8 0
4 years ago
If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation
Airida [17]

Answer:

59.077 kJ/mol.

Explanation:

  • From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

  • At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.

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