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qwelly [4]
3 years ago
5

If a temperature increase from 12.0 ∘c to 22.0 ∘c doubles the rate constant for a reaction, what is the value of the activation

barrier for the reaction?
Chemistry
1 answer:
Alex73 [517]3 years ago
6 0

Answer: 48,501 J/mol


Explanation:


1) Action barrier = activation energy = Ea


2) Data:


i) T₁ = 12°C = 12 + 273.15 K = 285.15K

ii) T₂ = 22°C = 22 + 273.15 K = 295.15 K

iii) rate constant = k: k₂ / k₁ = 2


iv) Ea = ?


3) Formula:


Arrhenius' law gives the relationship between the constant of reaction and the temperature:


k=Ae^{-\frac{Ea}{RT}}


4) Solution


By arranging the formula, you get:


㏑[k₂/k₁] =Ea/R [1/T₁ - 1/T₂]


Replace k₂ = 2k₁; T₁ = 285.15; and T₂ = 295.15


ln[2] = Ea/8.314 J/K mol × [1/285.15 - 1/295.15]K


Ea = ln [2] × 8.314 J/K mol / [1.18818×10⁻⁴K] = 48,501 J/mol





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Explanation:

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2 years ago
A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being
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Answer:

<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>

Explanation:

This question is about solubility.

Regarding solubility, the solutions may be classified as:

  • Unsaturated: the concentration is below the maximum concentration permited at the given temperature.

  • Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.

  • Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.

Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.

  • In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.

  • In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g  of water.

  • Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ

       115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O  ⇒ x =  57.5 g NaNO₃

  • <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>

<em />

3 0
3 years ago
Stearic acid (C18H36O2) is a fatty acid, a molecule with a long hydrocarbon chain and an organic acid group (COOH) at the end. I
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Answer:

There is 9671  Kj of heat released

Explanation:

<u>Step 1:</u> The balanced equation:

C18H36O2(s) + 26O2(g) --> 18CO2(g)+18H2O(g)

This means for 1 mole of C18H36O2 consumed there is 26 moles of O2 needed to produce 18 moles of CO2 and 18 moles of H2O.

<u>Step 2:</u> Calculate the heat of combustion

ΔH (combustion) = [18*(ΔHf of CO2) + 18*(ΔHf of H2O)] - [1*(ΔHf of C18H36O2) + 26*(ΔHf of O2)]

ΔH (combustion) = [18*(-394 kJ/mol) + 18*(-242 kJ/mol)] - [1*(-948 kJ/mol) + 26*(0 kJ/mol)]

ΔH (combustion) = [(-7092 kJ/mol) + (-4356 kJ/mol)] - (-948 kJ/mol)

= -10500 kJ/mol

ΔH (combustion) = heat released / number of moles of stearic acid

<u>Step 3:</u> Calculate moles of stearic acid

moles of stearic acid = mass / Molar mass of stearic acid

moles of stearic acid = 262g / 284.48 g/mole = 0.921 moles

<u>Step 4:</u> Calculate moles of oxygen

moles of O2 = 914.5 / 32g/mole

moles of O2 = 28.578125 moles

Stearic acid is the limiting reactant: it will <u>completely react</u>

There will react 26*0.921 mole = 23.946 mole of O2

This means there will remain 4.63 moles of O2

<u>Step 5:</u> Calculate heat released

q = (ΔH combustion) * (moles of stearic acid) = (-10500 kJ/mol) * (0.921 moles) = 9671 Kj

There is 9671  Kj of heat released

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Answer:

ΔT=-747,13°C

Explanation:

Sensible heat is<em> the amount of thermal energy that is required to change the temperature of an object</em>, the equation for calculating the heat change is  given by:

Q=msΔT

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To solve the problem, we clear ΔT of the equation and then replace our data:

Q=msΔT

ΔT=Q/ms

ΔT=\frac{-14900 J}{40,7g*0,49\frac{J}{gC} }=-747,13°C

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Thus, the change in temperature of the steel bar is -747,13°C, meaning that the temperature of the bar decreases.

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