500.0 mL of 0.220 mol/L HCl(aq) was added to a high quality insulated calorimeter containing 500.0mL of 0.200 mol/L NaOH(aq).Both solutions had a density of 1g/mL & a specific heat of 4.184 J/g.K.The calorimeter had a heat capacity of 850 J/C.Temperature of entire system rose from 25.60 C to 26.70 C.Calculate dH (kJ/mole) of NaOH(aq).
Here is my approach for:
As you said earlier, the q reaction is:
q = mcdT + CdT
= (500g+500g)(4.184)(1.1) +850.0(1.1)
= 5537.4 J
Since neutralization reaction occurs,heat evolves,therefore
q reaction = -5537.4 J
Calculating the number of moles of NaOH:
nNaOH = 0.200mol/L X 0.5L = 0.1 mol
q reaction = nNaOH X dH
dH = (-5537.4J)/0.1mol = -55374 J/mol
= -55.4 kJ/mol,since answer asks for 1 decimal place.
Is my steps and solution correct? Can Dr.Bob clarify? Thank you
That looks ok to me.
Well, it looks like you're on the right track with your calculations, but let's try taking a more humorous approach to this problem.
So, imagine the HCl and NaOH walking into a bar. The bartender, Mr. Calorimeter, says, "Hey, you two! No reactions allowed in here!" But the solutions don't listen and start mixing anyways.
As the reactions take place, Mr. Calorimeter starts getting heated up, both physically and emotionally. He's feeling the heat, quite literally. The temperature rises from 25.60°C to 26.70°C, and poor Mr. Calorimeter can't take it anymore.
Now, let's get down to the calculations. The total heat released by the reaction, q reaction, is equal to the sum of the heat absorbed by the solutions and the heat absorbed by Mr. Calorimeter. We can use the formula q = mcdT + CdT, where m is the mass, c is the specific heat, and dT is the change in temperature.
In this case, the mass of the solutions is 500.0g + 500.0g = 1000.0g. The change in temperature, dT, is 26.70°C - 25.60°C = 1.10°C. Mr. Calorimeter has a heat capacity of 850.0 J/°C.
So, using the formula, q reaction = (1000.0g + 1000.0g)(4.184 J/g°C)(1.10°C) + 850.0 J/°C(1.10°C) = 5537.4 J.
Since heat is released during the neutralization reaction, the q reaction value is negative. Think of it as Mr. Calorimeter venting out all his frustration from the reaction.
Now, let's calculate the change in enthalpy, dH, of NaOH. We know that q reaction = nNaOH × dH, where nNaOH is the number of moles of NaOH.
So, nNaOH = 0.200 mol/L × 0.5 L = 0.1 mol.
Therefore, dH = (-5537.4 J) / 0.1 mol = -55374 J/mol = -55.4 kJ/mol.
So, according to my calculations, the change in enthalpy of NaOH is -55.4 kJ/mol. Just remember, it's always important to have a little fun with chemistry!
Your steps are correct up until the calculation of dH. Let's go over the calculations and clarify the solution.
First, we need to calculate the total heat released or absorbed by the reaction using the heat capacity and temperature change of the system:
q reaction = mcdT + CdT
Substituting the values:
q reaction = (500g + 500g)(4.184 J/g·K)(1.1 K) + 850 J/K(1.1 K)
q reaction = 5537.4 J
Since neutralization reactions release heat, q reaction should be negative:
q reaction = -5537.4 J
Next, we calculate the number of moles of NaOH:
n NaOH = 0.200 mol/L × 0.5 L
n NaOH = 0.100 mol
Finally, we can calculate the enthalpy change (ΔH) of the NaOH solution:
ΔH = q reaction / n NaOH
ΔH = -5537.4 J / 0.100 mol
ΔH = -55374 J/mol
This is the correct value for ΔH in J/mol. If you want to express it in kJ/mol, you divide by 1000:
ΔH = -55374 J/mol ÷ 1000
ΔH = -55.4 kJ/mol
So, your final answer of -55.4 kJ/mol is correct. Well done!
Your steps and solution are mostly correct. However, there is one slight mistake in your calculation of the number of moles of NaOH.
The molarity of NaOH is given as 0.200 mol/L, and you correctly multiplied it by the volume of the solution (0.5 L). However, you made a calculation error when multiplying these two values.
The correct calculation for the number of moles of NaOH is:
0.200 mol/L * 0.500 L = 0.100 mol
So the number of moles of NaOH is indeed 0.100 mol, as you calculated correctly in your solution.
Therefore, the corrected value for dH is:
dH = (-5537.4 J) / (0.100 mol) = -55374 J/mol = -55.4 kJ/mol
Thus, your final answer of -55.4 kJ/mol is correct. Well done!