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jekas [21]
2 years ago
6

state the equation for the formation of ethane from its elements and ghen detwrmine the value for the standard enthalpy of forma

tion of ethane
Chemistry
1 answer:
RideAnS [48]2 years ago
3 0

The standard enthalpy of formation of ethane is determined as -192.5 kJ/mol.

<h3>Reaction for the formation of ethane</h3>

The reaction for the formation of ethane is given as;

2C + 3H₂ → C₂H₆

Where;

  • C is carbon (graphite)
  • H is hydrogen
  • C₂H₆ is ethane
<h3>Standard enthalpy of formation of ethane</h3>

The standard enthalpy of formation of ethane is calculated as follows;

ΔH = ΔH₁ + ΔH₂  + ΔH₃

where;

  • ΔH₁ is heat of formation of graphite = -286 kJ/mol
  • ΔH₂ is heat of formation of hydrogen = -393.5 kJ/mol
  • ΔH₃ is heat of formation of ethane = 1560 kJ/mol

2C + 3H₂ → C₂H₆

ΔH = 2(-286) + 3(-393.5) + 1(1560)

ΔH = -572 - 1180.5 + 1560

ΔH = -192.5 kJ/mol

Learn more about standard enthalpy here: brainly.com/question/14047927

#SPJ1

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What volume. In liters, of H2O(g) measured at STP is produced by the combustion of 15.63 g of natural gas (CH4) according to the
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Answer:

V = 43.95 L

Explanation:

Given data:

Mass of CH₄ decomposed = 15.63 g

Volume of H₂O produced at STP = ?

Solution:

Chemical equation:

CH₄ + 2O₂    →       2H₂O  + CO₂

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 15.63 g/ 16 g/mol

Number of moles = 0.98 mol

Now we will compare the moles of H₂O with CH₄.

                         CH₄              :              H₂O

                           1                 :                2

                        0.98             :            2×0.98 = 1.96 mol

Volume of hydrogen:

PV = nRT

1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K

V = 43.95atm.L / 1atm

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3 0
2 years ago
Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
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Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0
3 years ago
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