Calculate the enthalpy of combustion of ethyne (C2H2) from the data below: ΔHf (CO2) = -393kJ/mol , ΔHf (H2O) = -286kJ/mol and ΔHf (C2H2) = -229kJ/mol *

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

Solution:

The standard enthalpy of combustion is ΔH∘c. It is the heat evolved when 1 mol of a substance burns completely in oxygen at standard conditions.

The balanced chemical equation of combustion of ethyne (C2H2):

C2H2(g) + 2.5O2(g) → 2CO2(g) + H2O(l)

To calculate ΔH∘c from standard enthalpies of formation:

ΔH∘c = ∑ΔH∘f (products) − ∑ΔH∘f (reactants)

The enthalpy of formation for oxygen (O2) is zero.

ΔH∘f (C2H2) = +229 kJ/mol

ΔH∘f (CO2) = -393 kJ/mol

ΔH∘f (H2O) = -286 kJ/mol

Thus:

ΔH∘c = ΔH∘f (H2O) + 2×ΔH∘f (CO2) - ΔH∘f (C2H2)

ΔH∘c = (-286 kJ/mol) + 2×(-393 kJ/mol) - (+229 kJ/mol) = -1301 kJ/mol

Therefore, the enthalpy of combustion for ethyne is:

ΔH∘c = -1301 kJ/mol

Answer: The enthalpy of combustion of ethyne is -1301 kJ/mol.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
What is the percentage of hydrogen by mass in ethyne C2H2

The percentage by mass of Hydrogen in acetylene, C2H2 is 7.8%.

1 Answers 1 views
C + 2 H2O  CH4 + CO2 C + H2O  CO + H2 ∆H = +131.3 kJ CO + H2O  CO2 + H2 ∆H = -41.2 kJ CH4 + H2O  3 H2 + CO ∆H = +206.1 kJ

131.3+-41.2-206.1= 116kJ

1 Answers 1 views
the standard enthalpy of combustion of ethane, carbon, and hydrogen are -1559.7 kJ/mol, -393kJ/mol and -286kJ/mol respectively. Use Hess'ss law to find the enthalpy of formation of ethane

ΔHform(C2H6)=ΔHcomb(C2H6)-(2ΔHcomb(C)+3ΔHcomb(H2))=-1559.7-(2(-393)+3(-286))=84,3 kJ/mol

1 Answers 1 views
Define enthalpy of combustion. Calculate the enthalpy of combustion of methanol if the enthalpies of formation of CO2,H20 andCH3OH are respectively . -393.5kj/mol,-285.8kj/mol and -239.0kj/mol.

Answer: Enthalpy of combustion is the heat released during combustion of one mol amount of any substance by oxygen.You can calculate the combustion enthalpy of methanol using the enthalpy...

1 Answers 1 views
Assuming enthalpy of combustion of hydrogen at 273 K, –286 kJ and enthalpy of fusion of ice at the same temperature to be + 6.0 kJ, calculate enthalpy change during formation of 100 g of ice

Answer: H2 + 1/2O2 = H2O(l) ΔH (H2) =-286 kJ/mol ΔH H2O (s) = H2O(l) =6.0 kJ/mol n = m/M M (H2O) = 18 g/mol n (H2O) = 100 / 18...

1 Answers 1 views
What is the enthalpy change if 105 g of ethane is burned in a combustion reaction whose molar enthalpy of combustion is -1251 kJ/mol. *

n = m / MM (C2H6) = 30.07 g/moln (C2H6) = 105 / 30.07 = 3.5 moldelta H (C2H6) = - 1251 x 3.5 = -4367 kJ

1 Answers 1 views
Please pick the correct expression for the following equilibrium expression: CO (g) + H₂O (g) <--> CO₂ (g)+ H₂ (g) * [CO₂] [H₂] / [CO] [H₂O] [H₂] /[CO] [CO] [H₂O] / [CO₂] [H₂]

[CO₂] [H₂] / [CO] [H₂O] 

1 Answers 1 views
The combustion of 1.000g of benzene, C6H6 (l), in O2 (g) liberates 41.84kJ of heat and yields CO2(g) and H2O (l) . Write the thermochemical equation for the combustion of one mole of C6H6 (l) .

At first the balanced chemical equation should be written:2C6H6+15O2=12CO2+6H2O.It is seen that heat here should be written for 2 moles of benzene. So the given number of moles should be...

1 Answers 1 views
Calculate the enthalpy (∆H) for the following reaction: 1) H2O (g) + CO (g) = CO2 (g) + H2 (g) 2) C2H2 (g.) + 5 / 2O2 (g.) → 2CO2 (g.) + H2O (l.)

Species Delta H C2H2 (g )= +226.7 CO2 (g )= -393.5 H20 (g) = -241.6CO(g) = -99 kJ/mol.1. ∆H= -393.5 -(-99 + (-241.6))= -52,9 kJ/mol.2. ∆H= 2 x -393.5 +...

1 Answers 1 views
Calculate the entropy (∆S) for the following reaction: 1) H2O (g) + CO (g) = CO2 (g) + H2 (g) 2) C2H2 (g.) + 5 / 2O2 (g.) → 2CO2 (g.) + H2O (l.)

1. [1ΔSf(H2 (g)) + 1ΔSf(CO2 (g))] - [1ΔSf(H2O (g)) + 1ΔSf(CO (g))][1(130.59) + 1(213.68)] - [1(188.72) + 1(197.9)] = -42.35 J/K-42.35 J/K     (decrease in entropy)2. C2H2 + 5/2 O2 -> 2CO2...

1 Answers 1 views
X