Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
[tex](1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)[/tex]
The expression used will be:
[tex]Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})][/tex]
where,
[tex]Q[/tex] = heat released for the reaction = ?
m = mass of benzene = 94.4 g
[tex]c_{p,s}[/tex] = specific heat of solid benzene = [tex]1.51J/g^oC=1.51J/g.K[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid benzene = [tex]1.73J/g^oC=1.73J/g.K[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = [tex]-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g[/tex]
Now put all the given values in the above expression, we get:
[tex]Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K][/tex]
[tex]Q=-29427.312J=-29.4kJ[/tex]
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
Answer:
Explanation:
benzene boiling point = [tex]80^oC = 353.1K[/tex]
benzene freezing point = [tex]5.42^oC = 278.42K[/tex]
No of moles of benzene = [tex]\frac{94.48}{78.11} = 1.2095[/tex]
Energy released,
[tex]q = m*G*\delta T\\\\q = 94.4 * 1.73 * (278.42-322)\\\\q = -7117J[/tex]
Energy released = [tex]1.2095 * (-9.8*10^3)[/tex]
[tex]= -11853.1J[/tex]
Energy released,
[tex]q = n*C*\delta T\\\\q = 1.2095*118.4*(278.42-205)\\\\q = -10511.23J[/tex]
Total heat released
[tex]= -7117 -11853.1 -10511.23\\\\= -29481.3J\\\\= -29.48 KJ[/tex]
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