Respuesta :
Answer: The enthalpy change for the given process is -42.3 kJ/mol
Explanation:
The processes involved in the given problem are:
[tex]1.)H_2O(g)(100^oC)\rightarrow H_2O(l)(100^oC)\\2.)H_2O(l)(100^oC)\rightarrow H_2O(l)(80^oC)[/tex]
Pressure is taken as constant.
To calculate the amount of heat released at same temperature, we use the equation:
[tex]\Delta H=-L_{v}[/tex] ......(1)
where,
[tex]\Delta H[/tex] = enthalpy change
[tex]L_{v}[/tex] = latent heat of vaporization
To calculate the amount of heat absorbed at different temperature, we use the equation:
[tex]\Delta H=C_{p,m}\times (T_{2}-T_{1})[/tex] .......(2)
where,
[tex]\Delta H[/tex] = enthalpy change
[tex]C_{p,m}[/tex] = specific heat capacity of medium
[tex]T_2[/tex] = final temperature
[tex]T_1[/tex] = initial temperature
Calculating the enthalpy for each process:
- For process 1:
We are given:
[tex]L_v=40.79kJ/mol=40790J/mol[/tex]
Putting values in equation 1, we get:
[tex]\Delta H_1=-40790J/mol[/tex]
- For process 2:
We are given:
[tex]C_{p,l}=75.3J/mol.^oC\\T_1=100^oC\\T_2=80^oC[/tex]
Putting values in equation 2, we get:
[tex]\Delta H_2=75.3J/mol.^oC\times (80-(100))^oC\\\\\Delta H_2=-1506J/mol[/tex]
Enthalpy change of the reaction = [tex]\Delta H_1+\Delta H_2[/tex]
Enthalpy change of the reaction = [tex][-40790+-1506]J/mol=-42296J/mol=-42.3kJ/mol[/tex]
Hence, the enthalpy change for the given process is -42.3 kJ/mol
