Question

When 50.0 mL of water containing 0.50 mol HCl at 22.5°C are mixed with 50.0 mL of water containing 0.50 mol NaOH at 22.5°C in a calorimeter, the temperature of the solution increases to 26.0°C. How much heat (in kJ) was released by this reaction? Note: the specific heat of water (Cwater) is 4.18 J/(g•˚C) and the density of the solution is 1.00 g/mL.

Answer 1

Answer:

Explanation:

Hello,

At the specified conditions, to compute the released heat due to the mixing process, which also includes the effect of the reaction enthalpy, we take into account the following formula:

$H_{process}=H_{H_2O}+H_{rxn}$

All of the terms listed above are considered as changes. Thus, the enthalpy due to the water heating turns out into:

$H_{H_2O}=[(50.0mL+50mL)*\frac{1g}{1mL}]*4.18\frac{J}{mol^0C}*(26.0-22.5)^0C\\H_{H_2O}=146.3J=0.1463kJ$

Now, it is found that the enthalpy of the carried out reaction:

$NaOH+HCl-->NaCl+H_2O$

Has a value of -56.72kJ/mol but as 0.5 mol are involved, the total enthalpy turns out into -28.5kJ

Now, that reaction enthalpy is by far higher than the enthalpy due to the temperature increasing, thus, one concludes that the total released enthalpy is negative and has a value of:

$H_{process}=-(0.1463kJ+28.36kJ)=-28.5kJ$

Best regards.

Answer 2

Answer:

1.46kj

Explanation: