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### Topic: Calculating enthalpy change for two solution mixture  (Read 4585 times)

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#### KYR_Singularity

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##### Calculating enthalpy change for two solution mixture
« on: February 11, 2014, 03:13:53 PM »
25.0 cm3 of 2.00 mol dm-3 HCl(aq) was mixed with 25.0 cm3 of 2.00 mol dm-3 NaOH(aq). The temperature increased from 22.5°C to 34.5°C. Find the enthalpy change of reaction for the following equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l).

how do you work out the energy change in this reaction ?
please show all the work out  , thank you

#### zsinger

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##### Re: Calculating enthalpy change for two solution mixture
« Reply #1 on: February 11, 2014, 06:15:26 PM »
See:  Hess's Law.  Theres your hint….now lets see some work
Zack
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#### KYR_Singularity

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##### Re: Calculating enthalpy change for two solution mixture
« Reply #2 on: February 12, 2014, 05:22:10 PM »
HCL + NaOH -> NaCl + H2O
(25+25) x 4.18 x 12 = 2508 J
25 Volume reacted
25 x2 / 1000 = 0.05 mol
2508/0.05=50.16KJ/mol
^H = -50.16KJ/mol
is this right ? thanks for the tip anyway

#### Borek

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##### Re: Calculating enthalpy change for two solution mixture
« Reply #3 on: February 12, 2014, 05:41:18 PM »
Doesn't look bad, although you may want to take water produced to the total mass.

Some will consider it nitpicking, but it is almost a 2% difference.
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#### zsinger

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##### Re: Calculating enthalpy change for two solution mixture
« Reply #4 on: February 12, 2014, 07:23:38 PM »
Borek is correct.  Otherwise, you got it!
-Zack
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#### Benzene

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##### Re: Calculating enthalpy change for two solution mixture
« Reply #5 on: February 12, 2014, 08:18:19 PM »
In Hess's Law, it states that the sum δH products-δH reactants = δH reaction
Is the δH of the products and reactants calculated from the δH of formation? or combustion?

#### zsinger

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