Answer:
enthalpy of reaction = -53504.00 J/mol = -53.50 kJ/mol
Explanation:
The heat released in the reaction is the heat absorbed by the calorimeter. The heat absorbed by the calorimeter is given by the expresion:
Qcal = mx c x ΔT where
c = specific heat capacity = 4.18 J/gK
ΔT = the change in temperature = (28.2 - 25.5 ) K (in this case we do not need to convert ºC to K since what we need is the change)
Since the enthalpy of enthalpy of the reaction is the negative of Q we will finally need to convert this heat per mol HCl ( this is because we are told 0.025 mol HCl were reacted)
m = (50.0 + 50.0 mL) x 1g/mL = 100 g
Q = 100 g x 4.18 J/Kmol x 3.2 K = 1337.6 J
V HCl = 50.0 mL /1000 mL/L x 0.5 M = 0.025 mol
enthalpy = - 1337.6 J / 0.025 mol = -53504.00 J/mol = -53.50 kJ/mol
Answer:
[tex]\Delta _rH=- 53.5kJ/mol[/tex]
Explanation:
Hello,
In this case, by cause of the temperature change it is possible to compute that reaction's enthalpy of reaction as shown below:
[tex]\Delta _rH=mC\Delta T[/tex]
In such a way, by knowing that the mass results from adding both solutions' volumes, we obtain:
[tex]\Delta _rH=(50.0+50.0)mL*\frac{1g}{1mL}*4.18\frac{J}{g*K}*(301.35-298.15)K \\\Delta _rH=1337.6J[/tex]
Now, the moles of hydrochloric acid are computed as follows:
[tex]n_{HCl}=0.05L*0.5\frac{molHCl}{L}=0.025molHCl[/tex]
Finally, the enthalpy of reaction per mole of acid turns out:
[tex]\Delta _rH=-\frac{1337.6J}{0.025mol}*\frac{1kJ}{1000J} =- 53.5kJ/mol[/tex]
Taking into account that it is negative since heat is released due to the temperature increase.
Best regards.
