[funboy]

an experiment takes place as follows

5.5g of NaOH is added to 200ml of H2O

the temperature of H2O increases 2C from the reaction

I need to calculate DeltaH

DeltaH = total mass of reaction solution X temperature change X specific heat of reaction sollution

I am given the specific heat of the reaction sollution to be .00418 (or 4.18 X 10^-3)
Temperature change is 2
mass of reaction solution 200ml of water = 200g

Therefore

DetaH = 200 X 2 X .00418
= 1.672

I dont know if this is correct or not.

Any help would be greatly appreciated

does reaction solution = 200ml + 5.5g of NaOH = 205.5g ??

Thanks in adavnce

Chris

[Dan]

does reaction solution = 200ml + 5.5g of NaOH = 205.5g ??

Yes, mass(solvent) + mass(solute) = mass(solution), provided there is no release of gas etc. there is nowhere else for the mass to go.

[funboy]

When NaOH (5.5g)  was reacted with water(200ml) I noticed a 2C increase in temperature

Net Ionic Equation = NaOH(s) --> Na+(aq) + OH-(aq)

Temp change = 2 C
Mass of Solution = 5.5g + 200g = 205.5g
Heat Change (kJ) = specific heat X change in temp = 8.36 x 10^-3 kJ
Amount of NaOH (moles) = mass of NaOH/molar mass = 5.5g/40g = 1.38 X 10^-1 mol
Average H (kJ/mol NaOH) = heat change / amount of NaOH moles  = 6.1 x 10^-2 kJ/mol
DeltaH = m X DeltaT X Q
   = 205.5 X 2 X (4.18X10^-3)

Is any of this correct??
Should Temp Change be a negative because this is an exothermic reaction??

Thanks again