# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on July 03, 2021, 08:37:05 AM

Title: Enthalpy of reaction using one equation and its enthalpy
Post by: Win,odd Dhamnekar on July 03, 2021, 08:37:05 AM
How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?

HCl(aq) + NaOH (aq) :rarrow: NaCl(aq) + H2O(l) ΔT = -58 kJ

If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g°C, how much will the temperature increase? What assumption did you make in your calculation?

Reaction given produces 58 kJ  of heat with 1 mol of NaOH. We have 0.150 M of NaOH (limiting reagent) so it will produce 0.150 M × -58 kJ/1.0 M= -8.7 kJ of heat.
We know that q=cm∆T, substituting the values we have , in the equation -8.7 kJ = 4.19 J/g℃ × 300 g × ∆T, ⇒ ∆T= 6.9°C.

Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: Orcio_Dojek on July 03, 2021, 11:51:13 AM
No, it is wrong.
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: Win,odd Dhamnekar on July 03, 2021, 12:17:43 PM
Would you tell me  where i am wrong?
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: Orcio_Dojek on July 03, 2021, 12:38:03 PM
HCl is limitting agent (not NaOH) + emitted heat is calculated in wrong way.
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: mjc123 on July 03, 2021, 01:26:01 PM
OP, Do you understand what the symbol M means?
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: Win,odd Dhamnekar on July 03, 2021, 02:59:59 PM
Please read the following, interpret it and draw out the appropriate inference.
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: mjc123 on July 03, 2021, 05:17:03 PM
The appropriate inference is - no, you don't understand it.

(First of all, to dispose of what may be just a careless error, the unit of moles is not g/mol. It is moles. g/mol is the unit of molar mass.)

The symbol M DOES NOT mean moles. It means moles per litre. It is a unit of concentration, not amount. You do NOT have 0.250 moles of HCl. You have 100 mL of a 0.250 mol/L solution of HCl. How many moles is that?
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: Win,odd Dhamnekar on July 04, 2021, 12:14:45 AM
Thanks for finding out my mistake in my chemistry working.

Now, my rectified answer: We have 0.025 M of HCl (limiting reagent), so it will produce 0.025 M × -58 kJ/1 mol = -1.45 kJ of heat.
Substituting the known values in the heat equation q=cmΔT =4.19 J/g℃ × 300 g × ∆T=-1.45 kJ,

we get ΔT= -1.15℃

In my opinion, now this answer is correct.
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: mjc123 on July 04, 2021, 06:52:19 AM
The numerical calculation is correct. However:

1. You do not have 0.025 M of HCl. You have 0.025 moles. Reread my previous post. M DOES NOT MEAN MOLES!!!

2. A negative ΔH means heat is given out, and (unless there is efficient heat transfer to the surroundings) the temperature rises. A ΔH of -1.45 kJ means q, the heat entering the system, is +1.45 kJ. The original question asked "how much does the temperature increase?" It should be obvious that a negative ΔT is wrong.
Title: Re: Enthalpy of reaction using one equation and its enthalpy
Post by: Win,odd Dhamnekar on July 04, 2021, 09:37:13 AM
Yes, you are correct. ΔT =1.15°C is positive. Thanks again for pointing out where i was wrong in my chemistry workings.