Question
Question: At \(5 \times {10^5}\) bar pressure, the density of diamond and graphite are \[3g/cc\] and \[2g/cc\]...
At 5×105 bar pressure, the density of diamond and graphite are 3g/cc and 2g/cc respectively. At a certain temperature “T”. find the value of ΔU−ΔH for the conversion of 1 mole of graphite to 1 mole of the diamond at temperature” T”.
A.100KJ/mol
B. 50KJ/mol
C. -100KJ/mol
D. None of these
Solution
Enthalpy is a thermodynamic quantity and is often used by scientists to measure the energy changes in the chemical reactions and it is defined as the sum of internal energy of the system (U) and product of pressure (P) and Volume (V).
Complete step by step answer:
Since the diamond and graphite are the allotropes of carbon. Hence the weight of 1mole of graphite and diamond is 12 grams. As we know the relation between change in internal energy and change in enthalpy is given as:
ΔH=ΔU+PΔV (1)
Here, the change in enthalpy is represented by ΔH and the change in internal energy is represented by ΔU and the Pressure is represented by P. The change is volume is given as:
ΔV(Vgraphite−Vdiamond)
Now we will find the volume of diamond and volume of graphite, As we know Density=VolumeMass
Given that the density of diamond=3g/cc and the density of graphite is 2g/cc. On putting the value of density in the above formula we get:
3=volume of diamond12
⇒volume of diamond=312=4cm3
Similarly
2=volume of graphite12
⇒volume of graphite=212=6cm3
Now to calculate the change is volume, we will find the change in (ΔV) using the below formula:ΔV=Vgraphite−Vdiamond
⇒ΔV=6−4=2cm3
On putting the value of ΔVIn equation 1, we get:
ΔH=ΔU+5×105bar×2×10−3lt/mol
⇒ΔH=ΔU+103
⇒ΔH−ΔU=103
Taking negative sign common, the above equation becomes:
⇒ΔU−ΔH=−103bar Ltr
As we know 1 bar ltr=100J
Then −103bar ltr=−1000×100J/mol
Or
103bar ltr=100kJ/mol
Hence the correct answer is option (A).
Note:
Both graphite and diamonds are made of carbon and are known as allotropes of carbon, i.e. the arrangement of carbon atoms in space is different. The atoms of carbon in a diamond are arranged tetrahedrally and for the case of graphite, they are arranged in an infinite array.