2004 question 8 A

Could anyone help?

## 2004 - Written Paper - Q4(a)

### Isla from Banff

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

xx

hope that helps you out slightly

xx

### ad absurdum

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 cm

E

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 + H

We have 20cm

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 H = -54kJ mol

Hope that helps

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 cm

^{3}mass of water = 40g = 0.04kgE

_{h}= c m T = 4.18 * 0.04 * 6.5 = 1.0868 kJThe 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 + H

_{2}OWe have 20cm

^{3}1 mol per litre NaOH, this is the same as 0.02 mole of NaOHWe 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 H = -54kJ mol

^{-1}Hope that helps