Question

For the change, ${{\text{C}}_{{\text{diamond}}}} \to + {{\text{C}}_{{\text{graphite}}}}$; $\Delta H = - 1.89$kJ, if $6$g of diamond and $6$g of graphite are separately burnt to yield ${\text{C}}{{\text{O}}_{\text{2}}}$ the heat liberated in the first case is-
A.Less than in the second case by $1.89$ kJ
B.Less than in the second case by $11.34$ kJ
C.Less than in the second case by $14.34$ kJ
D.More than in the second case by $0.945$ kJ

Answer 1

Hint: First we will write the reaction of burning of the diamond allotrope and graphite allotrope separately. Then we will assume the heat liberated in the first case is $ - \Delta {H_1}$ and the heat liberated in the second case is $ - \Delta {H_2}$. Then we will subtract the equations obtained to get the given equation. So the enthalpy change will be subtracted in the same manner. Now, we know the value of enthalpy change of the given reaction so we will put its value in the equation obtained. Now we also see that the number of moles become half in the reaction of combustion of the allotropes so the given enthalpy change will also be halved. Then solve the equation to get the answer.

Complete step by step answer:
Given, for the change ${{\text{C}}_{{\text{diamond}}}} \to + {{\text{C}}_{{\text{graphite}}}}$ ,
The enthalpy change=$\Delta H = - 1.89$kJ
This is the reaction for $12g$ of diamond as the mass number of carbon=$12$ g
Now if $6$g of diamond and $6$g of graphite are separately burnt to yield ${\text{C}}{{\text{O}}_{\text{2}}}$ then we have to find the heat liberated when $6$g of diamond is burned to yield${\text{C}}{{\text{O}}_{\text{2}}}$.
The reaction of burning of diamond to produce ${\text{C}}{{\text{O}}_{\text{2}}}$is given as-
${C_{diamond}} + {O_2} \to C{O_2}$ --- (i)
Let the heat liberated in this reaction be $ - \Delta {H_1}$
Now the reaction of burning of graphite to produce ${\text{C}}{{\text{O}}_{\text{2}}}$is given as-
${C_{graphite}} + {O_2} \to C{O_2}$ --- (ii)
Let the heat liberated in this reaction be $ - \Delta {H_2}$
On subtracting eq. (ii) from (i) we get-
${{\text{C}}_{{\text{diamond}}}} \to + {{\text{C}}_{{\text{graphite}}}}$-- (iii)
Then the enthalpy change is given as-
$ \Rightarrow$ $ - \Delta {H_1} - \left( { - \Delta {H_2}} \right) = \Delta H$
We know the enthalpy change of eq. (iii) so we can write-
$ \Rightarrow $ $ - \Delta {H_1} - \left( { - \Delta {H_2}} \right) = - 1.89{\text{ kJ}}$
On solving, we get-
$ \Rightarrow $ $ - \Delta {H_1} + \Delta {H_2} = - 1.89{\text{ kJ}}$
On multiplying negative sign, we get
$ \Rightarrow $ $\Delta {H_1} - \Delta {H_2} = 1.89{\text{ kJ}}$--- (iv)
In this reaction only $6g$ of diamond is burned so the enthalpy change $\Delta H$ will become half as the moles of carbon in diamond form become half.
So, we get-
$ \Rightarrow$ $ \Delta {H_1} - \Delta {H_2} = \dfrac{{1.89}}{2}{\text{ kJ}}$
On solving, we get-
$ \Rightarrow$ $ \Delta {H_1} - \Delta {H_2} = 0.945{\text{ kJ}}$
Then we can write-
$ \Rightarrow$ $ \Delta {H_1} = 0.945{\text{ kJ + }}\Delta {H_2}$
This means the heat liberated in the first case or eq. (i) is more than the heat liberated in the second case by $0.945$ kJ.

The correct answer is option D.

Note:
Diamond and Graphite are allotropes of carbon as they are made up of carbon but the arrangement of carbon is different in them. The difference between them is-
Diamond has a crystalline structure while graphite has layered structure.
In diamond, the geometry is tetrahedral but in graphite the geometry is planar.
Diamond is extremely hard and transparent while graphite is soft and black in color.