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Tamiku [17]
3 years ago
14

Use Hess's Law to determine the enthalpy change (∆H) for the reaction: ClF + F2 → ClF3 Given: 2ClF + O2 → Cl2O + F2O. ∆H=167.4kJ

2ClF3 + 2O2 →Cl2O + 3F2O. ∆H=341.4kJ 2F2 + O2 →2F2O. ∆H=-43.4kJ
Chemistry
1 answer:
emmasim [6.3K]3 years ago
3 0

Answer:

The enthalpy change (∆H) for the reaction is -108.7 kJ

Explanation:

Hess's law can be stated as: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps. Then, Hess's Law states that the enthalpy of one reaction can be achieved by algebraically adding the enthalpies of other reactions.

So,  to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them.

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

2 ClF + O₂ → Cl₂O + F₂O ∆H=167.4kJ

Cl₂O + 3 F₂O → 2 ClF₃ + 2 O₂ ∆H= -341.4kJ  

The previous equation must be inverted, and the enthalpy value is also inverted, that is, the sign is changed.

2 F₂ + O₂ →2 F₂O ∆H=-43.4kJ

Reactants and products are added or canceled, taking into account that certain substances sometimes appear as a reagent and others as a product, so they are totally eliminated (there is nothing left of them anywhere in the reaction, if the same amount in reagents and products) or partially (this substance remains, in less quantity, only on one side), obtaining:

2 ClF + 2 F₂ → 2 ClF₃

Then, as all the reactants and products have a stoichiometric coefficient of 2, dividend by that number is obtained:

ClF + F₂ → ClF₃

Adding the enthalpies algebraically, and dividing by 2, because to get the "data" reaction you had to multiply by two, you get:

ΔH= [167.4 kJ - 341.4 kJ - 43.3 kJ]÷2

ΔH= -108.7 kJ

<u><em>The enthalpy change (∆H) for the reaction is -108.7 kJ</em></u>

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An enclosed vessel contains 2.5g of 9b nitrogen and 13.3g of chlorine at s.T.P. Of What will be the partial pressure of the Il n
kow [346]

Answer:

0.535 atm

Explanation:

Since the volume of the tank is constant, we use Gay- Lussac's law to find the pressure at 180°C.

So, P₁/T₁ = P₂/T₂ where P₁ = pressure at S.T.P = 1 atm, T₁ = temperature at S.T.P = 273.15 K, P₂ = pressure of gas at 180 °C and T₂ = 180 °C = 273.15 + 180 K = 453.15 K

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Substituting the values of the variables into the equation, we have

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P₂ = 1 atm × 453.15 K/273.15 K

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P₂ = 1.66 atm

We now need to find the total number of moles of each gas present

number of moles of nitrogen = mass of nitrogen, m/molar mass of nitrogen molecule M

n = m/M

m = 2.5 g and M = 2 × atomic mass of nitrogen (since it is diatomic) = 2 × 14 g/mol = 28 g/mol

So, n = 2.5 g/28 g/mol

n = 0.089 mol

number of moles of chlorine, n' = mass of chlorine, m'/molar mass of chlorine molecule M'

n' = m'/M'

m' = 13.3 g and M = 2 × atomic mass of chlorine (since it is diatomic) = 2 × 35.5 g/mol = 71 g/mol

So, n' = 13.3 g/71 g/mol

n' = 0.187 mol

So, the total number of moles of gas present is n" = n + n' = 0.089 mol + 0.187 mol = 0.276 mol

So, the partial pressure due to nitrogen gas, P = mole fraction of nitrogen × pressure of gas at 180 °C

P = n/n" × P₂

P = 0.089 mol/0.276 mol × 1.66 atm

P = 0.322 × 1.66 atm

P = 0.535 atm

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Rubber is an insulator, so heat has a hard time passing through (Ex. Using a rubber grip on a cast iron pan.

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