Hess Law

samiabuvla

CuSO4 (s) + 5H2O (l) CuSO4.5H2O (s)
For this reaction , how do you calculate enthalpy change

Reply 1

TypicalNerd

Original post

by samiabuvla

CuSO4 (s) + 5H2O (l) CuSO4.5H2O (s)
For this reaction , how do you calculate enthalpy change

What data have they given you?

Reply 2

samiabuvla

Original post

by TypicalNerd

What data have they given you?

I was just asking in general I don’t have any data

Reply 3

TypicalNerd

Original post

by samiabuvla

I was just asking in general I don’t have any data

Okay then, I presume you will need to work out solution enthalpies for CuSO4 (s) and CuSO4.5H2O (s), since both CuSO4 (s) and CuSO4.5H2O (s) dissolve in water to give [cu(h2o)6]^2+ (aq) and SO4^2- (aq).

Once you have these, you can set up a Hess cycle (which I’ll draw on paper later).

Of course, to obtain the relevant enthalpy changes, you can use the A level method where known amounts of each copper salt dissolving in some known amount of water of a known initial temperature in an insulated container.

You can plot a temperature-time graph and extrapolate Δθ from it, which you can substitute into the equation Q = mcΔθ to get the thermal energy (in J, where c is the specific heat capacity of water, 4.18 J/°C/mol, m is the mass of water and Δθ is the temperature rise - it will be positive for exothermic reactions and negative for endothermic reactions).

After which point, convert Q to kJ by dividing the result by 1000 and then using the formula ΔH = -Q/n (where n is the number of moles of copper salt dissolved) gives you the required value of ΔH.

(edited 10 months ago)

Reply 4

charco

Study Forum Helper

This very reaction was addressed in the attached video:

About 9m 08s
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