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Mice21 [21]
3 years ago
15

The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th

e formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone.
ClO(g)+O3(g)â¶Cl(g)+2O2(g) ÎHârxn=â122.8 kJ
2O3(g)â¶3O2(g) ÎHârxn=â285.3 kJ
O3(g)+Cl(g)â¶ClO(g)+O2(g) ÎHârxn= ?
Chemistry
1 answer:
OLga [1]3 years ago
4 0

Answer:

ΔH = -162.5 kJ.

Explanation:

Hello.

In this case, we first rearrange the reactions:

ClO(g) + O₃(g) ⇒ Cl(g) + 2O₂(g);  ΔH =-122.8 kJ

2O₃(g) ⇒ 3O₂(g);  ΔH=-285.3 kJ

O₃(g) + Cl(g) ⇒ ClO(g) + O₂(g);  ΔH= ?

Thus, we are going to use the Hess law, as an strategy to rearrange the known chemical reactions and thereby compute the enthalpy of reaction of the unknown one.

1. The first reaction must be inverted in order to obtain chlorine as a reactant in the third one, therefore, the enthalpy of reaction becomes positive:

Cl(g) + 2O₂(g) ⇒ ClO(g) + O₃(g);   ΔH = 122.8 kJ

2. Second reaction remains the same:

2O₃(g) ⇒ 3O₂(g);  ΔH=-285.3 kJ

Then, we add them to obtain:

Cl(g) + 2O₂(g) + 2O₃(g) ⇒ ClO(g) + O₃(g) + 3O₂(g)

Whereas we can subtract both oxygen and ozone to obtain the third one:

O₃(g) + Cl(g) ⇒ ClO(g) + O₂(g)

Therefore, the enthalpy of reaction turns out:

ΔH = 122.8 kJ + (-285.3 kJ )

ΔH = -162.5 kJ.

Best regards.

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giving the incomplete equation 2N2O5 (g) which set of products completes and balances the incomplete equarion
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Answer:

The complete question is as follows

Given the incomplete equation: 2 N2O5(g) ==>  Which set of products completes and balances the incomplete equation?

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C)4 NO2(g) + O2(g)

D)4 NO(g) + SO2(g)

The correct option is C) 4NO2(g) + O2(g)

Explanation:

Note that the products should be NO2 and O2 since the reactant is entirely made up of N and O. option A is not correct as hydrogen cannot emerge as a product in this reaction. Matter can never be created or be destroyed bu can only change in a chemical reaction. Option D is not also correct for the same reason.

Option B is not correct since it did not balance the number of atoms of O and N in the reactant side of the equation.

The option C) 4NO2(g) + O2(g) is therefore the right option since it balances both the elements and the number of atoms of the elements present.

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
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