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Maksim231197 [3]

2 years ago

12

Calculate the Lattice Energy of KCl(s) given the following data using the Born-Haber cycle: ΔHsublimation K = 79.2 kJ/mol IE1 K

= 418.7 kJ/mol Bond EnergyCl–Cl = 242.8 kJ/mol EACl = 348 kJ/mol ΔHKCl(s) = –435.7 kJ/mol

1 answer:

DanielleElmas [232]2 years ago

3 0

Answer:

Explanation:

The following equation relates to Born-Haber cycle

$\triangle H_f = S + \frac{1}{2} B + IE_M - EA_X+ U_L$

Where

$\triangle H_f$ is enthalpy of formation

S is enthalpy of sublimation

B is bond enthalpy

$IE_M$ is ionisation enthalpy of metal

$EA_X$ is electron affinity of non metal atom

$U_L$ is lattice energy

Substituting the given values we have

-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L

$U_L$ = - 707 KJ / mol

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ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

The combustion of propane occurs when propane interacts with oxygen gas to produce water and carbon dioxide in the following rea

gtnhenbr [62]

Answer:

20 atm

Explanation:

<u>1) Data:</u>

a) n = 2 moles

b) T = 373 K

c) V = 2.5 liter

d) P = ?

<u>2) Chemical principles and formula</u>

You need to calculate the pressure of the propane gas in the mixture before reacting. So, you can apply the partial pressure principle which states that each gas exerts a pressure as if it occupies the entire volume.

Thus, you just have to use the ideal gas equation: PV = nRT

<u>3) Solution:</u>

PV = nRT ⇒ P = nRT / V

P = 2 mol × 0.08206 atm-liter /K-mol × 373K / 2.5 liter = 24.5 atm

Since the number of moles are reported with one significant figure, you must round your answer to one significant figure, and that is 20 atm (20 is closer to 24.5 than to 30).

5 0

2 years ago

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) How many grams of HC

SVETLANKA909090 [29]

Answer:

The correct answer is is option B

b. 93.3 g

Explanation:

SEE COMPLETE QUESTION BELOW

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)

How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?

a. 7.30 g

b. 93.3 g

c. 146 g

d. 150 g

e. 196 g

CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION

7 0

2 years ago

A 0.286-g sample of gas occupies 125 ml at 60. cm of hg and 25°

irga5000 [103]

Using the Equation: PV=nRT
Where P is the pressure 60 cmHg or 600 mmHg or 600/760= 0.789 atm
V is the volume 125 ml or 0.125 L, n is the number of moles, R is a constant 0.082057, and T is temperature 25 °C or 298 K;
Therefore:
0.789 × 0.125 = n × 0.082057 × 298
n = 0.0987/24.45
= 0.004036 mol
0.004036 mole has a mass of 0.286 g
Hence; 1 mole has a mass of 0.286/0.004036
= 70.8 g /mol
Therefore the molar mass of the gas is 71 g/mol (2 sfg)

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2 years ago

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Veronika [31]

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2 years ago