HCl is a corrosive colourless gas that dissolves readily in water.
Aqueous HCL reacts with NaOH to form water and NaCl. In a simple
calorimeter, a 100.00 mL sample of 0.355 mol/L HCl(aq) is mixed
with 50.00 mL of excess NaOH(aq). During the reaction, there is a
rise in temperature by 4.200 °C. Calculate the molar enthalpy change
for the above reaction.
I'm really confused on this question, I don't know where to start.
This is what I did:
(I'm using the heat capacity of water, I don't know if it was the right choice, can you tell explain please?)
Assuming 1 ml = 1g
Q =mc∆t
=(150.00g)(4.184 J/g c)(4.200 c)
= 2635.92 J
= 2.63592 KJ
molar enthalpy change = 2.63592 KJ/ 0.355 mol/L
Then I'm stuck, what am I suppose to do ?
You have almost all of it correct.
Q = mc*dT. yes
Yes, use 150 mL because that's the total volume. Yes, use heat capacity for water. What was it that changed temperature. It was the water so that's the heat capacity to use. USUALLY, the problem says "in a coffee cup calorimeter" or "assume there is no loss of heat to the surroundings" but in this case the problem doesn't even list a calorimeter. Also note that sometimes a calorimeter is used AND the problem gives the heat capacity of the calorimeter. In that case Q = mc*dT (for the water) + Ccal*dT (for the calorimeter. In this case Ccal*dT is not needed.
Next: so you have calculated Q for the heat of the reaction and listed it as 2.64 kJ. That's for how many moles? You list M = molarity. You want MOLES. moles = M x L = 0.355 M x 0.100 L = 0.0355 mols.
So 2.64 J/0.0355 mol = ? kJ/mole
Follow up with any question if you still don't get it.
I have a question, isn't it suppose to be 0.150 L instead of the 0.100 L because " 100.00 mL sample of 0.355 mol/L HCl(aq) is mixed
with 50.00 mL of excess NaOH(aq). " or is it just 0.100 L?
"I have a question, isn't it suppose to be 0.150 L instead of the 0.100 L because " 100.00 mL sample of 0.355 mol/L HCl(aq) is mixed
with 50.00 mL of excess NaOH(aq). " or is it just 0.100 L?"
I'm glad you asked. No, it is just 0.100 L. Why? In this case the HCl is the limiting reatgent (it says the 50 mL NaOH IS AN EXCESS); therefore, if we asked ourselves how many moles NaOH will react (probably more moles NaOH were there) the answer is the same number of moles HCl we had initially. Moles HCl initially is M x L = 0.355 x 0.1 M = 0.0355. It makes no difference the number of moles NaOH present BECAUSE ONLY 0.0355 moles NaOH will react with the 0.0355 mols HCl we had. And of course it doesn't matter how much volume NaOH we added as long as we had enough NaOH to be greater so that we reacted with ALL of the HCl. That extra 50 mL NaOH must be added to te water that was heated because ALL of the water was heated but the amount of water holding that 0.0355 moles NaOH doesn't change the amount of moles reacting. Hope this helps.
To calculate the molar enthalpy change for the reaction, you're on the right track with using the heat capacity of water to determine the heat released or absorbed during the reaction. Here is a step-by-step guide to help you through the calculation:
Step 1: Calculate the moles of HCl reacted.
You have a 100.00 mL sample of 0.355 mol/L HCl(aq). Use the formula:
moles = concentration (mol/L) × volume (L)
moles = 0.355 mol/L × 0.100 L
moles = 0.0355 mol
Step 2: Calculate the heat released or absorbed during the reaction.
The reaction between HCl and NaOH is exothermic, meaning it releases heat. Use the equation:
q = m × c × ΔT
q = heat released or absorbed (J)
m = mass of the solution (g)
c = specific heat capacity of water = 4.184 J/g°C (since you're assuming 1 mL = 1 g)
ΔT = temperature change (°C)
First, calculate the mass of the solution:
mass = volume × density
Since you assumed 1 mL = 1 g, the mass would be:
mass = 150.00 g (100.00 mL + 50.00 mL)
Now, use the heat capacity equation to find q:
q = 150.00 g × 4.184 J/g°C × 4.200 °C
q = 2512.08 J (or 2.51208 kJ)
Step 3: Calculate the molar enthalpy change.
The molar enthalpy change (ΔH) is the heat released or absorbed per mole of HCl reacted. Divide q by the moles of HCl:
ΔH = q / moles
ΔH = 2512.08 J / 0.0355 mol
ΔH = 70,728.9 J/mol (or 70.7289 kJ/mol)
So, the molar enthalpy change for the reaction is approximately 70.7289 kJ/mol.