Which is true about a chemical reaction

A solution contains two isomers, n-propyl alcohol and isopropyl alcohol, at 25°C. The total vapor pressure is 38.6 torr. What ar

Leona [35]

Answer:

Mole fraction of alcohols in liquid phase $x_1=0.2727\& x_2=0.7273$ .

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$ .

Explanation:

The total vapor pressure of the solution = p =38.6 Torr

Partial vapor pressure of the n-propyl alcohol = $p^{o}_1=21.0 Torr$

Partial vapor pressure of the isopropyl alcohol = $p^{o}_2=45.2 Torr$

$p=x_1\times p^{o}_1+x_2\times p^{o}_2$ (Raoult's Law)

$p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2$

$38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr$

$x_1=0.2727$

$x_2=1-0.2727=0.7273$

$x_1\& x_2$ is mole fraction in liquid phase.

Mole fraction of components in vapor phase $y_1\& y_2$

$p_1=y_1\times p$ (Dalton's law of partial pressure)

$y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}$

$y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468$

$y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}$

$y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516$

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$

5 0

3 years ago

Rewrite each equation below with the delta H value included with either the reactants or the products and identify the reaction

yanalaym [24]

A) 2H₂(g) + O₂(g) → 2H₂O(l) + 285.83 kJ
Exothermic

B) 2Mg + O₂ → 2MgO + 1200kJ
Exothermic

5 0

3 years ago

I REALLY NEED HELP can someone help me please I’ll mark you brainliest! :)

Yanka [14]

Answer:

A

Explanation:

5 0

3 years ago

Describe how a constant volume calorimeter can be used to measure the heat transferred from a burning piece of coal to cool wate

Vlad1618 [11]

Answer:

A Calorimeter is a device used to measure the amount of heat involved in a chemical or physical process.

Explanation:

Ex- when an exothermic reaction ( accompanied by or requiring absorption of heat)occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution increasing its temperature.

- When an endothermic reaction (Heat evolving reaction) occurs, the heat required is absorbed from the thermal energy of the solution, decreasing its temperature.

A burning piece of coal which is releasing heat due to its combustion and is transferred to water.

Mark as Brainliest ;)

7 0

3 years ago

Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Answer: The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

For hydrogen gas:

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

For water:

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

6 0

4 years ago