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Iteru [2.4K]
3 years ago
11

A certain reaction with an activation energy of 155 kJ/mol was run at 495 K and again at 515 K . What is the ratio of f at the h

igher temperature to f at the lower temperature
Chemistry
1 answer:
wolverine [178]3 years ago
6 0

Answer:

4.32 is the ratio of f at the higher temperature to f at the lower temperature

Explanation:

Using the sum of Arrhenius equation you can obtain:

ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)

<em>Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)</em>

Replacing:

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

<em>Where 2 represents the state with the higher temperature and 1 the lower temperature.</em>

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

ln (f₂/f₁) = 1.4626

f₂/f₁ = 4.32

<h3>4.32 is the ratio of f at the higher temperature to f at the lower temperature</h3>

<em />

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Answer:

3.6124 m/kg

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Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us  35.4 g of H3PO4 and 100 g of solution to work with.

To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.

Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:

35.4 g x 1 mol / 97.9952 g = 0.3612422 mol

Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.

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Explanation :

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According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

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(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

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\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

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