# Molar enthalpy change/calorimetry

• Chemistry
Homework Statement:
Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.

NaOH(s) → Na+(aq) + OH-(aq) + .... kJ
50.0 ml of water is in a cylinder (temperature is 25 degrees C). 1.10 grams of NaOH is in a foam cup. When the water is poured into the foam cup the temperature changes to 37.6. What is the molar enthalpy change in kilojoules per mole of NaOH? I got -44.07 kJ

Can you please help me with the other two reactions?

Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an aqueous solution of sodium chloride.

NaOH(s) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq) + .... kJ
Reaction 2 is the same as reaction 1 except that 49.0 ml is used in place of water. (49.0 ml of 0.5 M HCl is in the cylinder and its temperature is 25.00 C. The final temperature is 37.62 C.
Calculate the molar enthalpy change of NaOH?

Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an aqueous solution of sodium chloride.

Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq) + ..... kJ

24.0 ml of 1.00 M hydrochloric acid solution is in a foam cup. An equal volume of 1.00 M sodium hydroxide solution is in a cylinder. The temperature of both solutions is 25 degrees C.
Sodium hydroxide solution is poured into the foam cup and the temperature obtained is 31.67 degrees C.
Calculate the energy released (molar enthalpy change) in kilojoules per mole of sodium hydroxide used.
Relevant Equations:
Q=mc(delta T)
deltha H= -Q
n=M/m
specific heat capacity of water is: 4.18 j/g.C
I am struggling to find the molar enthalpy change of reactions 2 and 3. I am confused what is the surrounding for reaction 2 and 3? what is the mass of the surrounding? I really need help with this.
For reaction one I got: molar enthalpy change of NaOH is -44.07.

## Answers and Replies

mjc123
Science Advisor
Homework Helper
Your answer to 1 is wrong. Please show your working.
For 2, what is the temperature change?

Your answer to 1 is wrong. Please show your working.
For 2, what is the temperature change?
reaction 1: Q=mc (delta T) -> Q= (50g)(4.18 J/g. C)(5.8 C)= 1212.2 J
delta H= -Q -> delta H= -1212.2 J or -1.2122 kJ
n (NaOH)= m/M
n= (1.10 g) /( 39.997 g/mol)
n= 0.0275 mol
molar enthalpy of NaOH= enthalpy change/ number of moles
= -1.2122 kJ/0.0275
= -44.076 kJ/ mol of NaOH

Please tell me where I made a mistake

For reaction 2 delta T is: 37.62 C - 25.00 C = 12.62 C

Your answer to 1 is wrong. Please show your working.
For 2, what is the temperature change?
Could you tell me the correct answer to 1?

Borek
Mentor
Hint: reaction heat heats up the solution, not just solvent.

mjc123
Science Advisor
Homework Helper
For reaction 1, the temperature change is 12.6 K, not 5.8. It ought not to be the same for reaction 2. Did you mistakenly quote the final temperature for reaction 2 under reaction 1, and forget it for 2?

For reaction 1, the temperature change is 12.6 K, not 5.8. It ought not to be the same for reaction 2. Did you mistakenly quote the final temperature for reaction 2 under reaction 1, and forget it for 2?
my bad!
Reaction 1 -> Ti= 25 Tf= 30.8 delta T=5.8
Reaction 2 -> Ti=25 Tf=37.62 delta T=12.62
Raection 3 -> Ti=25 Tf=31.67 delta T= 6.67

Last edited:
Hint: reaction heat heats up the solution, not just solvent.
How would I use this hint to solve this question? can you explain more

For reaction 1, the temperature change is 12.6 K, not 5.8. It ought not to be the same for reaction 2. Did you mistakenly quote the final temperature for reaction 2 under reaction 1, and forget it for 2?
I put the correct temperature change in my last reply. Now, that temperature change is 5.8 for reaction one, is my answer to it correct?

mjc123
Science Advisor
Homework Helper
Your calculation is correct. Borek's point is that the total mass of the solution is 51.1 g, not 50 g. However, the solution heat capacity will be a bit different from that of pure water, in a way that is not easy to predict, so you're probably best using water values, 50 g x 4.18 J/g/K. For dilute solutions the error will be small.

Your calculation is correct. Borek's point is that the total mass of the solution is 51.1 g, not 50 g. However, the solution heat capacity will be a bit different from that of pure water, in a way that is not easy to predict, so you're probably best using water values, 50 g x 4.18 J/g/K. For dilute solutions the error will be small.
I once read that when you have a solid that is added to a solution like water, you cant add the masses (reaction 1 and 2). Unless you have a solution, then you can add the solutions together (reaction 3).
Am I right??