Solveeit Logo
Login

Question

Question: \( CC{{l}_{2}}{{F}_{2}} \) cooled \( 1.25 \) gm sample at constant atmospheric pressure of \( 1 \) a...

CCl2F2CC{{l}_{2}}{{F}_{2}} cooled 1.251.25 gm sample at constant atmospheric pressure of 11 atm from 320K320K to 293K.293K. During cooling the sample volume decreased from 274274 to 248248 mL. Calculate ΔH\Delta H for the CCl2F2CC{{l}_{2}}{{F}_{2}} for this process. (CP=80.7 J mol−1K−1)\left( {{C}_{P}}=80.7\text{ }J\text{ }mo{{l}^{-1}}{{K}^{-1}} \right)
(A) 22.51 J22.51\text{ }J
(B) −19.83 J-19.83\text{ }J
(C) 19.88 J19.88\text{ }J
(D) −22.51 J-22.51\text{ }J

Explanation

Solution

We know that the measurement of enthalpy and internal energy change is carried out by an experimental approach called calorimetric. These techniques are established on thermometric procedures which are done in a vessel known as a calorimeter that is submerged in a known liquid volume.

Complete answer:
The heat energy, which is evolved or absorbed during a chemical reaction progression, is called enthalpy. It is represented by H, and the letter H indicates the energy amount. Enthalpy change is given by ΔH\Delta H where the delta symbol shows the change and its unit is joules or kilojoules. It can be said that the sum of internal energies of the system is enthalpy. The reason is that a change in internal energy takes place at the time of chemical reaction, and this change is calculated as enthalpy. As ΔH\Delta H is expressed as the flow of heat at constant pressure, calculations done using a calorimeter of constant-pressure (a system utilized to measure changes in enthalpy during chemical reactions at constant pressure) gives out the direct value of delta H enthalpy.
Given mass sample =1.25g=1.25g
Mw=CCl2F2=[(12×2)+(35.5×2)+(19×2)]=133g{{M}_{w}}=CC{{l}_{2}}{{F}_{2}}=\left[ \left( 12\times 2 \right)+\left( 35.5\times 2 \right)+\left( 19\times 2 \right) \right]=133g
⇒C=80.7133=0.606 J g−1\Rightarrow C=\dfrac{80.7}{133}=0.606\text{ }J\text{ }{{g}^{-1}}
Here, we have ΔT=(−320+290)=−30K\Delta T=\left( -320+290 \right)=-30K
ΔH=m×CP×ΔT=1.25×0.6060×(−30)=−22.51 J\Delta H=m\times {{C}_{P}}\times \Delta T=1.25\times 0.6060\times \left( -30 \right)=-22.51\text{ }J
Therefore, the correct answer is option D.

Note:
Remember that a certain amount of a substance is taken in the cup of Chloroform linked with electrical wires for creating an arc to ignite combustion. After that, the vessel is tightly closed, and pressure is exerted by putting in excess oxygen. The vessel or bomb is submerged in water, in the calorimeter’s inner volume.