Question

Question #a3360

Answer 1

`DeltaH=-54.5(kJ)/(mol)`

Explanation:

To calculate the amount of energy involved when water cools down from `105^@C` to `-2^@C`, we will consider the phase transitions that the water go through:

`H_2O(g, 105^@C)->H_2O(g, 100^@C)" " "q_1`

`H_2O(g, 100^@C)->H_2O(l, 100^@C)" " "q_2`

`H_2O(l, 100^@C)->H_2O(l, 0^@C)" " " "q_3`

`H_2O(l, 0^@C)->H_2O(s, 0^@C)" " " "q_4`

`H_2O(s, 0^@C)->H_2O(s, -2^@C)" " "q_5`

The overall equation will be:

`H_2O(g, 105^@C)->H_2O(s, -2^@C)" " "q_t`

where, `q_t=q_1+q_2+q_3+q_4+q_5`

Now we will need to calculate each heat evolved for each equation:

To find, `q_1` we will consider the heat involved using:

`q=mxxsxxDeltaT`

where, `q` is the amount of heat, `m` is the mass of water, `s=2.0J/(""^@C*g)` is the specific heat capacity of steam and `DeltaT=T_f-T_i` is the change on temperature.

`=>DeltaT=100-105=-5^@C`

Thus, `q_1=4.0cancel(g)xx2.0J/(cancel(""^@C)*cancel(g))xx(-5)cancel(""^@C)=-40.0J`

To find `q_2`, we will use the `DeltaH_(cond)=-40.7(kJ)/(mol)` of water:

The number of mole of water is: `n=m/(MM)=(4.0cancel(g))/(18cancel(g)/(mol))=0.22mol`

Thus, `q_2=(-40.7kJ)/(1cancel(mol))xx0.22cancel(mol)=-9.0kJ=-9000.0J`

To find `q_3`, we will consider the heat involved when converting water liquid from `100^@C` to `0^@C` using the specific heat capacity of water liquid (`s=4.18J/(""^@C*g)` is the specific heat capacity of water liquid):

`q=mxxsxxDeltaT`

Thus, `q_3=4.0cancel(g)xx4.18J/(cancel(""^@C)*cancel(g))xx(0-100)cancel(""^@C)=-1672.0J`

To find `q_4`, we will use the `DeltaH_("freezing")=-6.02(kJ)/(mol)` of water:

Thus, `q_4=(-6.02kJ)/(1cancel(mol))xx0.22cancel(mol)=-1.32kJ=-1320J`

To find `q_5`, we will consider the heat involved when converting water liquid from `0^@C` to `-2^@C` using the specific heat capacity of ice (`s=2.03J/(""^@C*g)` is the specific heat capacity of water liquid):

`q=mxxsxxDeltaT`

Thus, `q_5=4.0cancel(g)xx2.03J/(cancel(""^@C)*cancel(g))xx(-2-0)cancel(""^@C)=-16.0J`

Therefore, `q_t=q_1+q_2+q_3+q_4+q_5=-40.0J+(-9000.0J)+(-1672.0J)+(-1320J)+(-16.0J)=-12048=-12.0kJ`

`=>DeltaH=q/n=(-12.0)/(0.22mol)=-54.5(kJ)/(mol)`

Thermochemistry | Enthalpy and Calorimetry.