Question

What is the ratio of enthalpy yield on combustion of hydrogen atoms to steam to the yield on combustion of an equal mass of hydrogen molecules to steam?
Given: ${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O$, $\Delta H=-240KJ$, Bond energy (H-H) = 436 KJ.
A. 0.80:1
B. 1:0.80
C. 1.80:1
D. 2.80:1

Answer 1

Hint: Enthalpy is also known by the name heat content which can be described as a process takes place at constant pressure then heat is absorbed or released is equal to the enthalpy change. Enthalpy is derived from the Greek word with the meaning warming.

Complete step by step answer:
- Bond energy is also known by the name average bond enthalpy and this term can be explained as the amount of energy needed to break one chemical bond.
- Combustion enthalpy is defined as the enthalpy change when one mole of a compound is completely burnt in oxygen with all reactants and products in their standard state under standard conditions.
${{H}_{2}}+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O$, $\Delta H = -240KJ$, Bond energy (H-H) = 436 KJ.
- Combustion enthalpy of this reaction can be calculated by:
$\Delta H=B.E(H-H)+\dfrac{B.E(O=O)}{2}-2B.E(O-H)$ = -240 KJ and bond energy of H-H = 436 KJ given now put the values in the equation
$\dfrac{B.E(O=O)}{2}-2B.E(O-H)=-242-436=-678KJ$
- Combustion enthalpy for hydrogen molecules can be calculated as:
$2H+\dfrac{1}{2}{{O}_{2}}\to {{H}_{2}}O$
$\Delta H=2(0)+\dfrac{B.E(O=O)}{2}-2B.E(O-H)=-678KJ$
Now their ratio will be $\dfrac{678}{240}=2.8:1$
The correct answer is option “D” .

Note: Bond energy of chemical bonds is directly proportional to the stability of the bond which describes that greater the bond energy of a given chemical bond between the atoms then more the stability of that chemical bond.