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Topic: Enthalpy of reaction using one equation and its enthalpy  (Read 829 times)

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Offline Win,odd Dhamnekar

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Enthalpy of reaction using one equation and its enthalpy
« 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?

My answer:
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.

Is this answer correct?

« Last Edit: July 03, 2021, 12:22:44 PM by Win,odd Dhamnekar »
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Offline Orcio_Dojek

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #1 on: July 03, 2021, 11:51:13 AM »
No, it is wrong.

Offline Win,odd Dhamnekar

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #2 on: July 03, 2021, 12:17:43 PM »
Would you tell me  where i am wrong?
Any science consists of the following process.
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5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

Offline Orcio_Dojek

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #3 on: July 03, 2021, 12:38:03 PM »
HCl is limitting agent (not NaOH) + emitted heat is calculated in wrong way.

Offline mjc123

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #4 on: July 03, 2021, 01:26:01 PM »
OP, Do you understand what the symbol M means?

Offline Win,odd Dhamnekar

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #5 on: July 03, 2021, 02:59:59 PM »
Please read the following, interpret it and draw out the appropriate inference.
Any science consists of the following process.
 1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

Offline mjc123

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #6 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?

Offline Win,odd Dhamnekar

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #7 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.
Any science consists of the following process.
 1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

Offline mjc123

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #8 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.

Offline Win,odd Dhamnekar

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Re: Enthalpy of reaction using one equation and its enthalpy
« Reply #9 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.
Any science consists of the following process.
 1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

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