Please can u help me answer both questions a and b thanks

Anna [14]

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

<h3 /><h3>FIRST STEP</h3>

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

<h3>SECOND STEP</h3>

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

<h3>THIRD STEP</h3>

Finally, you need 4 moles of HF (g) on the product side. The first equation has 2 moles of HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

<h3>SUMMARY</h3>

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g) ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g) ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g) ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g) ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

Learn more about molar enthalpy:

7 0

2 years ago

Which of the following lists elements from lowest to highest atomic number

vitfil [10]

Lowest is Hydrogen highest is <span>Beryllium
-HOPE THIS HELPED </span>

8 0

3 years ago

What ion (cation or anion and charge) do the halogen group (17) form to fulfill the octet

Dafna11 [192]

Answer:

option B= anion (-1)

Explanation:

The halogens are present in the P- block of periodic table. There are thirty five elements present in p-block. The halogens are present in group seventeen. Their are seven electrons are present in valance sheet. All halogens require one electron to complete the octet and form anion carry -1 charge.

Halogens include following elements:

Fluorine, chlorine, bromine, iodine and astatine.

The halogens mostly form single bond with the other elements by gaining one electron and -1 charge. These are most reactive non-metals.

Properties of some halogen elements:

Fluorine:

1. It is yellow in color.

2. It is flammable gas.

3. It is highly corrosive.

4. Fluorine has pungent smell.

5. It's reactions with all other elements are vigorous except neon, oxygen, krypton and helium.

Chlorine:

1. It is greenish-yellow irritating gas.

2. Its melting point is 172.2 K.

3. Its boiling point is 238.6 K.

4. It is disinfectant and can kill the bacteria.

5. It is also used in manufacturing of paper, paints and textile industries.

Iodine:

1. It is present in solid form.

2. It is crystalline in nature.

3. It is very corrosive and has pungent odor.

4. It can not react with oxygen and nitrogen.

6 0

3 years ago

Vamos para playa oh te rajas lol

Anuta_ua [19.1K]

Answer:

let's go to the beach or u will relax lol

6 0

3 years ago

A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of

pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = <em>0.20</em>

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I hope it helps!

3 0

3 years ago