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DaniilM [7]
3 years ago
5

Using the bond energies provided below, calculate ?H° for the reaction

Chemistry
1 answer:
shusha [124]3 years ago
8 0

Answer:

ΔHºrxn =  - 440 kJ

Explanation:

What we need to remember in this question is that given the bond energies for a reaction, the enthalpy change , ΔHº , is given by the expression:

ΔHºrxn = ∑ energies of the bonds broken   - ∑ energies of the bonds formed.

where s  ∑ is  summatory.

The reaction is CH₄(g) + 4 Cl₂(g) ⇒ CCl₄(g) + 4 HCl(g)

So lets make the inventory of the bond broken and bonds formed

Bonds Broken:

4 C-H = 4 mol (413 kJ/mol ) = 1652 kJ

4 Cl-Cl = 4 mol ( 243 kJ/mol = 972 kJ/mol

Bond Formed:

4 C-Cl = 4 mol (339 kJ/mol ) = 1356 kJ

4 H-Cl = 4 mol (427 kJ/mol ) = 1708 kJ

Now we can calculate  ΔHºrxn :

ΔHºrxn = ∑ energies of the bonds broken   - ∑ energies of the bonds formed

ΔHºrxn = ( 1652 + 972 ) kJ   - ( 1356 + 1708 ) kJ

            = 2624 kj  - 3064 kJ

            = - 440 kJ

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a solution of unknown molecular substance is prepared by dissolving 0.50g of the unknown in 8.0g of benzene. the solution freeze
spayn [35]

Answer:

THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.

Explanation:

Mass of the unknown substance = 0.50 g

Freezing point of the solution = 3.9 °C

Freezing point of pure benzene = 5.5 °C

Freezing point dissociation constant Kf = 5.12°C/m

First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.

Change in temperature = 5.5 -3.9 = 1.6 °C

Next is to calculate the number of moles or molarity of the compound that dissolved.

Using the formula:

Δt = i Kf m

Assume i = 1

So,

1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene

x = 1.6 * 0.008 / 5.12

x = 0.0128 / 5.12

x = 0.0025 moles.

Next is to calculate the molar mass using the formula, molarity = mass / molar mass

Molar mass = mass / molarity

Molar mass = 0.50 g /0.0025

Molar mass = 200 g/mol

Hence, the molar mass of the unknown compound is 200 g/mol

6 0
3 years ago
Which reactions have a positive Δrxn? A(g)+B(g)⟶C(g) 2A(g)+2B(g)⟶5C(g) A(s)+B(s)⟶C(g) 2A(g)+2B(g)⟶3C(g)
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Answer:

A(g)+B(g)⟶C(g) 2A(g)+2B(g)⟶5C(g)

A(s)+B(s)⟶C(g) 2A(g)+2B(g)⟶3C(g):

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3 years ago
2. Classify the following solutions as acidic, basic, or neutral at 25OC.
PilotLPTM [1.2K]

Answer: a) pH = 13.00 : basic

b) [H_3O^+]=1.0\times 10^{-12}: basic

c) pOH = 5.00 : basic

d) [OH^-]=1.0\times 10^{-9}: acidic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H_3O^+]

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

a) pH = 13.00

As pH is more than 7, the solution is basic.

b)  [H_3O^+]=1.0\times 10^{-12}

Putting in the values:

pH=-\log[1.0\times 10^{-12}]

pH=12

As pH is more than 7, the solution is basic.

c) pOH = 5.00

pH+pOH=14.0

pH=14.0-5.00=9.00

As pH is more than 7, the solution is basic.

d) [OH^-]=1.0\times 10^{-9}

Putting in the values:

pOH=-\log[1.0\times 10^{-9}]

pOH=9.00

pH+pOH=14.0

pH=14.0-9.00=5.00

As pH is less than 7, the solution is acidic.

3 0
3 years ago
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