Using the following thermochemical data, what is the change in enthalpy for the following reaction:3H2(g) + 2C(s) + ½O2(g) → C2H5OH(l)
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l), ΔH = –1367 kJ/mol
C(s)+O2(g)→CO2(g), ΔH = –393.5 kJ
H2(g)+½O2(g)→H2O(l), ΔH = –285.8 kJ

a. -277.6 kJ/mol
b. -194.7 kJ/mol
c. 194.7 kJ/mol
d. 486 kJ/mol

Answers

Answer 1
Answer:

Answer:

The answer is a. -277.6 kJ/mol

Explanation:

We have to arrange and sum the given equations to obtain the desired reaction as follows:

3 H₂O(l) + 2 CO₂(g) → 3 O₂(g) + C₂H₅OH(l)       ΔH1= (-1) x (-1367 KJ/mol)

2 C(s) + 2 O₂(g)→ 2 CO₂(g)                    ΔH2= 2 x (-393.5 KJ/mol)

3 H₂(g) + 3/2 O₂(g) → 3 H₂O(l)               ΔH3= 3 x (-285.8 KJ/mol)

In the first equation (ΔH1) , we multiply the enthalphy by (-1) because we use the reverse reaction. In the second and third equations (ΔH2 and ΔH3), we multiply by 2 and 3, respectively. Once we sum and cancel the terms that are repeated in both sides of the reactions, we obtain:

3H₂(g) + 2C(s) + ½O₂(g) → C₂H₅OH(l)

The total enthalphy is:

ΔH= ΔH1 + ΔH2 + ΔH3

ΔH= (-1) x (-1367 KJ/mol) + 2 x (-393.5 KJ/mol) + 3 x (-285.8 KJ/mol)

ΔH= -277.4 KJ/mol

The nearest answer between the given options is option a (-277.6 KJ/mol)

Answer 2
Answer: The answer to this question is A

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Glycerol is a syrupy liquid often used in cosmetics and soaps. A 3.25-L sample of pure glycerol has a mass of 4.10 X 10^3 g. What is the density of glycerol in g/cm^3?

Answers

Answer: The density of glycerol will be 1.26cm^3

Explanation:

Density is defined as the mass contained per unit volume.

Density=(mass)/(Volume)

Given : Mass of glycerol= 4.10* 10^3 g

Volume of bromine =  3.25 L= 3250cm^3  (1L=1000cm^3)

Putting in the values we get:

Density=(4.10* 10^3g)/(3250cm^3)=1.26g/cm^3

Thus density of glycerol will be 1.26cm^3

The density of glycerol used in cosmetics and soaps is 1.26 g/cm³.

FURTHER EXPLANATION

Density is the measure of the compactness of a material. It is the mass per given volume. This can be expressed mathematically as:

Density = (mass)/(volume)  (equation 1)

To solve the problem, we first sort the given in the problem:

mass = 4.10 × 10³ g

volume - 3.25 L

The problem is asking for the density in units of g/cm³ which means that we must first convert the volume from liters to cm³.

volume \ (in \ cm^3) = 3.25 \ L * (1000 \ mL)/(1 \ L) * (1 \ cm^3)/(1 \ mL)\nvolume \ (in \ cm^3) = 3250 \ cm^3

Now we can plug in the values of mass and volume into equation 1:

Density = (4.10 * 10^3 \ g)/(3250 \ cm^3)\n\nDensity = 1.2615 (g)/(cm^3)

Since the given in the problem both have 3 significant figures, the final answer must also have three significant digits only. Therefore,

\boxed {Density = 1.26 (g)/(cm^3)}

The density of glycerol is greater than 1, so when mixed with water it will sink.

LEARN MORE

  1. Learn more about Displacement Method brainly.com/question/11474085
  2. Learn more about Liquid Density brainly.com/question/11301218
  3. Learn more about Dimensional Analysis brainly.com/question/1557970

Keywords: Density, Mass, Volume, Unit Conversion

What does a 100 lb person weigh on Uranus

Answers

If you were 100 lb on Earth, you would be 92 lb on Uranus

To neutralize gastric juices in your stomach, antacids contain what

Answers

Antacid contains weak acid neutralizing substances like calcium carbonate and aluminum hydroxide. These compounds are slow acting as result does not provide fast relief from hyperacidity due to excess acids in the stomach.

Would you expect the water or sugar solution to have the most supercooling? WHY? PLEASE HELP

Answers

Answer:

The water solution

Explanation:

Super freezing occurs when energy is quickly withdrawn from a liquid and it cools below the solidification temperature. This phenomenon occurs due to the fact that the cooling is very fast and does not occur the formation of an initial "point" or crystal, which is generally formed to start the freezing. A small crystal formed will attract the molecules nearby, forming a large crystal, that is, the frozen liquid.

The sugar solution should be distilled so that Super Cooling occurs more efficiently than the water solution because the presence of the sugar would create nucleation points and allow the formation of ice crystals. The super cooled water solution turns very quickly into ice, much faster than the sugar solution. It is also possible to supercooled the water after the homogeneous nucleation point, in which case it eventually solidifies into a type of glass.

Supercooled water can be prompted to trun into ice by tends to make it more difficult to achieve low nucleation Trehalose and other sugar solutions at low.

What is the total number of electrons shared in a double covalent bond?(1) 1 (3) 3
(2) 2 (4) 4

Answers

The correct answer is option 4. In a double covalent bond, two pairs of electrons are bonded or there is a total of four electrons being shared in a bond. For example, we have oxygen. Two atoms of oxygen share two pairs of electrons to have a stable structure.

What is the limiting reactant if 4.0 g of Nh3 react with 8.0 g of oxygen?

Answers

Answer : The limiting reagent is O_2.

Solution : Given,

Mass of NH_3 = 4.0 g

Mass of O_2 = 8.0 g

Molar mass of NH_3 = 17 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of NH_3 and O_2.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}=(4.0g)/(17g/mole)=0.24moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(8.0g)/(32g/mole)=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

4NH_3+5O_2\rightarrow 4NO+6H_2O

From the balanced reaction we conclude that

As, 5 mole of O_2 react with 4 mole of NH_3

So, 0.25 moles of O_2 react with (0.25)/(5)* 4=0.20 moles of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is O_2.

NH3-The limiting reactant is the reactant that get completely used up in a reaction