Question

Bond enthalpies of C=O, C-H, H-O, O=O are 741, 414, 464, 498 kJ mol-1 respectively. calculate the enthalpy of formation of methane.

Answer 1

Answer:

Explanation:

Given data:

Bond energies:

C=O = 741 KJ/mol

C-H = 414 KJ/mol

H-O = 464 KJ/mol

O=O = 498 KJ/mol

Enthalpy of formation of methane = ?

Solution:

ΔH = Bond broken - bond formation

Chemical equation:

CO₂ + 4H₂  →  CH₄ +2H₂O

CO + 3H₂   →  CH₄ + H₂O

For first equation:

Bond broken:

C=O = 741 × 2 = 1482  KJ/mol

H-H = 436 × 4 = 1744  KJ/mol

              Total energy = 3226 KJ/mol

Bond formed:

C-H = 414 × 4 = 1656  KJ/mol

O-H = 464 × 4 = 1856  KJ/mol

     Total energy = 3512 KJ/mol

ΔH = 3226 - 3512  

ΔH = -286 Kj/mol

For second reaction:

Bond broken:

C≡O = 1072 KJ/mol

H-H = 436 × 3 = 1308 KJ/mol

              Total energy = 2380 KJ/mol

Bond formed:

C-H = 414 × 4 = 1656  KJ/mol

O-H = 464 × 2 = 928  KJ/mol

     Total energy = 2584 KJ/mol

ΔH = 2380 - 2584

ΔH = -204 Kj/mol

Answer 2

Given:

Bond enthalpies value:

C=O = 741 KJ/mol

C-H = 414 KJ/mol

H-O = 464 KJ/mol

O=O = 498KJ/mol

To find:

Enthalpy of formation of methane = ?

Formula to be used:

ΔH = Total energy required for breaking of bond - Total energy required for bond formation.

Calculation:

The chemical reaction involved in formation of Methane is as follows

$CO_2+4H_2\rightarrow CH_4+2H_2O$

To find the enthalpy of formation of methane, energy required for breaking as well as formation of bond for the above reaction is required.

Energy required for breaking of bond:

C=O = 741×2 = 1482KJ/mol

H-H = 436×4 = 1744 KJ/mol

Total energy required = 1482 + 1744 = 3226KJ/mol

Energy required for bond formation

C-H = 414×4 = 1656KJ/mol

O-H = 464×4 = 1856 KJ/mol

Total energy required = 1656+1856 = 3512KJ/mol

∴ΔH = Bond breaking energy - Bond formation energy

ΔH = 3226-3512

ΔH = -286KJ/mol

Conclusion:

The enthalpy of formation of methane is -286KJ/mol

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