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2004 - Written Paper - Q4(a) - HSN forum

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2004 - Written Paper - Q4(a)


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#1 will_789

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Posted 11 April 2006 - 01:53 PM

2004 question 8 A

Could anyone help?

#2 Isla from Banff

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Posted 13 April 2006 - 11:02 AM

I can't remember how to work it out but i have the answer which is -54kJ mol^-1

hope that helps you out slightly

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#3 ad absurdum

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Posted 13 April 2006 - 07:23 PM

First off calculate the enthalpy change for the reaction:
Average temp at the start = 19 degrees celsius
Temp at end = 25.5 degrees celsius
Change in temp = 6.5 degrees celsius
volume of water = 40 cm3 therefore.gif mass of water = 40g = 0.04kg

Eh = c m delta.gif T = 4.18 * 0.04 * 6.5 = 1.0868 kJ

The enthalpy of neutralistion is when an alkali is neutralised by an acid to form one mole of water. In this experiment we have:
NaOH + HCl --> NaCl + H2O

We have 20cm3 1 mol per litre NaOH, this is the same as 0.02 mole of NaOH
We also have 0.02 mole of HCl, for the same reason
From the equation we can see that they will react to form 0.02 mole of water

So the enthalpy change we have just worked out (1.0868kJ of heat given out) was for 0.02 mole
Since the enthalpy of neutralisation is the enthalpy change when an acid neutralises an alkali to form one mole of water, we have to predict how much heat is going to be released when one mole of water is formed
since 0.02 mole produces 1.0868kJ, 1 mole will release (1.0868/0.02)kJ = 54kJ

Since the heat is given out (you know this as the final solution is a higher temperature than the average of the two reacting solutions) the enthalpy change is negative, so delta.gif H = -54kJ mol-1

Hope that helps
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