Homework 1: Calculating Enthalpy Change from Bond Energies
Use the table below to answer the following questions.

Table 1 Average Bond Energies (kJ/mol)

Bond Energy
H-H 432
H-F 565
C-H 413
C-O 358
C=O Triple bond 1072
C-C 347

F-F 154
O-H 467
C=C 614
C=O 745
C=O (for CO₂(g)) 799
0-0 495

Calculate the enthalpy change from bond energies for each of these reactions:
1. H2(g) + F2(g) → 2 HF(g)

ΔΗ=

2. CH4(g) +202(g) → CO2(g) + 2H₂O (g)

ΔΗ =

3. 2H2(g) + O2(g) → 2H₂O(g)

ΔΗ=

4.2H₂O(g) 2H₂(g) + O₂(g)

ΔΗ =

5. CH4(g) + H₂O(g) →CO(g) + 3H₂(g)

ΔΗ=

Answers

Answer 1

From the question;

1) The enthalpy  is  544 kJ/mol

2) The enthalpy is -110 kJ/mol

3) The enthalpy is -425 kJ/mol

4) The enthalpy is  425 kJ/mol

What is the bond energy?

Bond energy, sometimes referred to as bond dissociation energy or bond strength, is the amount of energy needed to completely dissociate the bound atoms and break a chemical bond. It expresses the potency of the attraction forces that hold the atoms together and measures the stability of a chemical bond.

The enthalpy of the reaction is obtained from;

Enthalpy of reaction = Sum of bond energy of products - Sum of bond energy of reactants

1) 2(565) - [432 + 154]

= 544 kJ/mol

2) [2(799) + 2(467)] - [(4 * 413) + 2(495)]

(1598 + 934) - (1652 + 990)

2532 - 2642

= -110 kJ/mol

3) 2(467) - [2(432) + (495)]

934 - 1356

= -425 kJ/mol

4) [2(432) + (495)] - 2(467)

= 425 kJ/mol

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Related Questions

how to explain the energy released by the reaction in terms of the law of conservation of energy

Answers

Chemical reactions can either absorb or release energy depending on the energy requirements of the reaction. When the products have a lower energy content than the reactants, the reaction is exothermic and releases energy to the surrounding environment.

This energy can be in the form of heat, light, or sound depending on the nature of the reaction. The amount of energy released by a reaction can be calculated by measuring the enthalpy change of the reaction, which is the difference between the enthalpy of the products and the enthalpy of the reactants. The law of conservation of energy states that energy can neither be created nor destroyed, but can only be converted from one form to another. In the case of an exothermic reaction, the energy released by the reaction is equal to the energy absorbed by the surroundings, as energy is conserved. This means that the total energy of the system and surroundings remains constant before and after the reaction. For example, when hydrogen gas reacts with oxygen gas to form water, a large amount of energy is released in the form of heat and light. This is an exothermic reaction as the energy content of the water is lower than the energy content of the hydrogen and oxygen gases. According to the law of conservation of energy, the total energy of the system and surroundings remains constant, so the energy released by the reaction is equal to the energy absorbed by the surroundings. In conclusion, the energy released by a chemical reaction is explained by the law of conservation of energy, which states that energy is conserved and cannot be created or destroyed, only converted from one form to another. Therefore, in an exothermic reaction, the energy released by the reaction is equal to the energy absorbed by the surroundings, ensuring that the total energy of the system and surroundings remains constant.

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A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by:

ATP(aq) + H2O(l) <----> ADP(aq) + HPO42-(aq)
for which delta Gorxn = - 30.50 kJ/mol at 37.0 oC and pH 7.0. Calculate the value of delta Grxn in a biological cell at human body temperature in which the concentrations of ATP, ADP, and HPO42- are 5.6 mM, 0.3 mM, and 4.98 mM, respectively.

Report your answer in units of kJ/mol to 2 decimal places.

Answers

Alright fam, let's dive into this biochem stuff. It's all about energy, which is something we can all vibe with, right?

First up, we got the standard change in Gibbs free energy (ΔGº). That's like the default energy change happening when ATP breaks up into ADP and a phosphate ion. In this case, it's -30.50 kJ/mol at body temp (37.0 oC), at pH 7.0.

But things get interesting when we step into the real-world scenario, aka inside a human cell, where the ATP, ADP, and phosphate ion (HPO42-) concentrations aren't at "standard" levels (which are usually 1 Molar for each reactant and product).

Now, to calculate the Gibbs free energy (ΔG) under these conditions, we use a super handy formula:

ΔG = ΔGº + RT ln(Q)

where:

- R is the universal gas constant (8.314 J/mol*K), but for kJ/mol, we'll use 0.008314

- T is the temperature in Kelvin (we add 273.15 to the Celsius temperature to convert it, so 37.0 oC becomes 310.15 K)

- Q is the reaction quotient, which is products over reactants. For us, it's the concentrations of ADP and phosphate ion divided by the concentration of ATP. Keep in mind that water is not included 'cause its concentration is assumed to be constant in biological reactions.

So now let's run the numbers:

ΔG = -30.50 kJ/mol + (0.008314 kJ/mol*K * 310.15 K) * ln((0.3 mM * 4.98 mM) / 5.6 mM)

= -30.50 kJ/mol + (2.58 kJ/mol) * ln(1.49)

= -30.50 kJ/mol + 1.04 kJ/mol

And finally, we get:

ΔG = -29.46 kJ/mol

So that's it. In the cell, the Gibbs free energy change is -29.46 kJ/mol, a bit different from the "standard" conditions, thanks to the concentration levels in a human cell. The negative sign tells us the reaction is spontaneous, meaning it happens on its own without needing an energy push. So yeah, your cells are legit energy machines.

Hope that's clear, and remember, biochem is like the most epic game of tiny molecular Legos you could ever play!

Assignment Your Unde a professor in a University has Sent you an touration 6 his Inaugural lectore wate a letter to him, showing appreciation for him on halind gesture and Congratulating! his achievements So far​

Answers

In this letter, express gratitude to your uncle, a university professor, for his invitation and congratulate him on his achievements.

Here are the steps to be followed:

Start with a proper salutation: Dear [Uncle's Name],

Express gratitude and appreciation: Begin the letter by expressing your gratitude for your uncle's kind gesture in inviting you to his inaugural lecture. Show genuine appreciation for his thoughtfulness in including you in such an important event.

Congratulate your uncle on his achievements: Extend your heartfelt congratulations to your uncle for his accomplishments thus far. Acknowledge his hard work, dedication, and commitment to becoming a professor at the university. Highlight specific achievements or milestones that you find particularly impressive.

Share your excitement and anticipation: Express your excitement and anticipation about attending his inaugural lecture. Let your uncle know that you are looking forward to being a part of this significant moment in his career and witnessing his expertise in action.

Offer support and encouragement: Offer words of encouragement and support for your uncle's future endeavors. Let him know that you are proud of him and believe in his continued success. Encourage him to keep pursuing his passion and making a positive impact in his field.

Close the letter with warm regards: End the letter with a closing remark and warm regards. You can use phrases such as "Best regards," "With love," or "Sincerely," followed by your name.

Proofread and revise: Before finalizing the letter, review it for any errors or areas that may need improvement. Ensure that the tone is respectful, appreciative, and heartfelt.

Send the letter: Once you are satisfied with the letter, send it to your uncle either via traditional mail or email, depending on your preferred method of communication.

By following these steps, you can write a thoughtful and appreciative letter to your uncle, expressing your gratitude for his invitation and congratulating him on his achievements.

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Reaction of carbon with hydrogen gas produces propane gas (C3H8).
3 C(s) + 4 H₂(g) →→→ C3H8(g) A Hrxn = -105 kJ

Answer the following questions.

Is this reaction endothermic or exothermic?

How may grams of carbon is necessary to produce 1220kJ of heat?

Answers

Approximately 139.47 grams of carbon are necessary to produce 1220 kJ of heat in this reaction.

The reaction of carbon with hydrogen gas to produce propane gas is an exothermic reaction. This can be determined based on the given information that the enthalpy change of the reaction (ΔHrxn) is -105 kJ. A negative value for ΔHrxn indicates that the reaction releases heat energy to the surroundings, indicating an exothermic process.

To calculate the amount of carbon required to produce 1220 kJ of heat, we need to use the balanced equation and the stoichiometry of the reaction.

From the balanced equation:

3 moles of carbon (C) produce 1 mole of propane (C3H8)

ΔHrxn = -105 kJ (heat released per mole of propane)

We can use the molar ratio between carbon and propane to calculate the moles of carbon required:

3 moles of carbon → 1 mole of propane

Moles of carbon = Moles of propane x (3 moles of carbon / 1 mole of propane)

= 1220 kJ / (-105 kJ/mol)

≈ -11.62 moles (note the negative sign indicates heat released)

Since we cannot have a negative number of moles, we can ignore the sign and take the absolute value. So, we need approximately 11.62 moles of carbon.

To convert moles of carbon to grams, we need to know the molar mass of carbon, which is 12.01 g/mol.

Mass of carbon = Moles of carbon x Molar mass of carbon= 11.62 moles x 12.01 g/mol

≈ 139.47 g

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A vial of Ancef 1 g is reconstituted with 5 mL of normal saline to yield 125mg / m * L How many mL of the medication should be given if a patient is prescribed 250 mg of the medication?

Answers

Ancef (Cefazolin) is a first-generation cephalosporin antibiotic that is used to treat bacterial infections. Cefazolin is available in several formulations, including injectable, intravenous, and powder for injection.

A vial of Ancef 1 g is reconstituted with 5 mL of normal saline to yield 125mg / m * L. We need to determine how many milliliters of the medication should be given if a patient is prescribed 250 mg of the medication.To begin with, let us first calculate the concentration of the reconstituted solution using the given data.1 gram of Ancef (Cefazolin) = 1000 milligrams (mg)5 mL of normal saline = 5000 milligrams (mg)Therefore, the total volume of the reconstituted solution = 5 mL + the volume of Ancef (Cefazolin)1 g of Ancef (Cefazolin) = 125 mg/mL (Given)Therefore, the volume of Ancef (Cefazolin) = (250 mg)/(125 mg/mL) = 2 mLTherefore, the total volume of the reconstituted solution = 5 mL + 2 mL = 7 mLThus, the amount of medication that should be given to the patient is 2 mL.

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What two air masses form a STATIONARY front?

Captionless Image
maritime polar
continental polar
maritime tropical
continental tropical

Answers

A stationary front is a boundary between two different air masses that aren't moving relative to each other, but instead are stationary.An air mass is a massive body of air with uniform temperature and moisture characteristics. An air mass takes on the qualities of the area where it forms. As an air mass moves from one place to another, it carries its temperature and moisture content with it.

It can change temperature and humidity, but not as quickly as it can change location.Types of air massesThe air masses are categorized based on two criteria. The first is the place of formation, while the second is the kind of surface over which they move. There are four types of air masses based on the place of formation, and two types based on the surface over which they move, for a total of six different types. Maritime tropical (mT): Warm and moist air masses that originate over water.Maritime polar (mP): Cold and moist air masses that originate over water.Continental tropical (cT): Dry and hot air masses that form over land. Continental polar (cP): Dry and cold air masses that form over land.A stationary front can be created when a mass of air with a uniform temperature and moisture characteristic meets an opposing air mass with similar characteristics but moving in a different direction. When a cold front and warm front meet and neither front is powerful enough to move the other, a stationary front occurs. The result is a stationary front that separates the two opposing air masses, like continental polar and continental tropical.

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