Class Day/ Time Lz/ CHEM 160A Gen. Chem. 1 Lab-Final A I3 uation that you use an

Question

Class Day/ Time Lz/ CHEM 160A Gen. Chem. 1 Lab-Final A I3 uation that you use and SHOw prpriaificant igures, round appropriately, and include appropriate units. nd SHOW ALL WORK!l Credit may not be given you obtained your answer. Report values to the cach cquaintely showing how Write correet number of significan Paints WIL.L. HE DEDUCTED 6 Total point value for the final is 40i le wthout appropriately showingwsow UED for unit less answers, unless it is a unit less numl t value for the final is 40pts. number I. Calorimetry (12pts) a ICI was mixed with 9.50g of NaOII in a calorimeter, the temperature of Whens ncreased from 18.5°C to 29.10°c cnts) Calculate the heat released (in joules) by the reaction. Assume that the specific a. When 3.35g of HCI was m the solution increased fi . (ópts) Calcu heat of t calorimeter is totally perfect of the solution i the solution is the same as that of pure water (4.184 J/g-°C) and that the i. (@pts) Calculate the enthalpy of reaction in kJ/mol of HCI (MM- 36.5g/mol). iii. (2pts) This reaction is thermic

Explanation / Answer

You have a lot of missing data.

a.

I will assume that 3.35 g is the mass of a solution of HCl.

I will assume that 9.50 g is the mass of a solution of NaOH.

The total mass of the solution mixture is 12.65 g.

Use the calorimetric equation for a perfect calorimeter (that have heat capacity = 0)

Qreleased = mass x specific heat x dT

              = 12.85 g x 4.184 J/g.ºC x 10.6 ºC

              = 570 J

b.

Assume that the concentration of HCl was 10% w/w and that HCl was the limiting reagent.

3.35 g x 10/100 = 0.335 g

0.335g/36.5g/mol = 0.0092 mol HCl

Q released per 1 mol HCl was

570 J/0.0092 mol = 62 x103 J/mol = 62 kJ/mol

This is also the value of the enthalpy of reaction, but with a minus sign (because the reaction is exothermic)

dH = - 62 kJ/mol

…………………………………

Bonus:

For the neutralization of a strong acid with a strong base, the enthalpy of neutralization has almost the same value for all acid and bases. The explanation is based on the fact that the only reaction here is

H+    +   HO-     = H2O        dH = - 57…58 kJ/mol

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