500.0mL of 0.220 mol/L HCl(aq) was added to a high quality insulated calorimeter containing 500.0mL of 0.200mol/L NaOH(aq).Both solutions had a density of 1.000g/mL & a specific heat of 4.184 J/g.K. The calorimeter had a heat capacity of 850.0 J/degree C. The temperature of the entire system changes from 25.6 to 26.7C. Calculate dH in kJ/mole for NaOH.
I know the initial part of this problem.
q = mcdT
m of HCI = density X volume
= 1 g/ml X 500ml= 500 g
m of NaoH = 1g/mL X 500ml = 500 g
q = (500+500)(4.184)(1.1)
= 4602.4 J
Hope my approach is correct to this point.My problem is I don't know how to relate the heat capacity of calorimeter to the problem and how to find dH of NaOH eventually.
Need help. Thank you.
The calorimeter absorbs heat also; the way to handle that is to add it to the number you've calculated for the rest of the system.
4602.4 J + (Ccal x delta T) = 850 J/C * 1.1 = ?
Ok,then how do I find the dH of NaOH? Do I need to use the total heat of the system and divide it by the mass of NaOH to get its dH ? I'm still confused
Thank you
Sorry what I meant was dividing the total heat with mole of NaOH ,not mass.
Your approach for calculating the heat transferred (q) is correct. Now let's proceed to calculate the change in enthalpy (ΔH) for NaOH.
In this experiment, the calorimeter is acting as a heat sink, absorbing the heat released by the reaction between HCl and NaOH. The heat capacity of the calorimeter is the amount of heat energy needed to raise the temperature of the calorimeter by 1 degree Celsius. We need to consider the heat absorbed by the calorimeter to account for the complete energy balance.
The equation for the heat transferred in the reaction can be written as:
q = q_solution + q_calorimeter
Where q_solution represents the heat transferred by the HCl and NaOH solutions, and q_calorimeter represents the heat absorbed by the calorimeter.
We already calculated q_solution as 4602.4 J.
To calculate q_calorimeter, we use the equation:
q_calorimeter = C_calorimeter * ΔT
Where C_calorimeter is the heat capacity of the calorimeter and ΔT is the change in temperature of the calorimeter.
Given:
C_calorimeter = 850.0 J/°C
ΔT = 26.7 °C - 25.6 °C = 1.1 °C
Therefore:
q_calorimeter = 850.0 J/°C * 1.1 °C
= 935 J
Now, to find the total heat transferred (q_total) in the reaction, we sum up q_solution and q_calorimeter:
q_total = q_solution + q_calorimeter
= 4602.4 J + 935 J
= 5537.4 J
Now we need to convert this to kJ:
q_total = 5537.4 J * (1 kJ / 1000 J)
= 5.5374 kJ
To calculate the enthalpy change (ΔH) per mole of NaOH, we need to consider the moles of NaOH in the reaction. The balanced equation for the reaction between HCl and NaOH is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
From the balanced equation, it is clear that the mole ratio of NaOH to HCl is 1:1. Therefore, we have 0.220 moles of NaOH in the reaction.
ΔH = q_total / moles of NaOH
= 5.5374 kJ / 0.220 mol
= 25.169 kJ/mol
Therefore, the enthalpy change (ΔH) for NaOH is 25.169 kJ/mol.