Thermochem problems! Please Help!?

Question

I am kind of having trouble with these problems...I get to a certain point then I'm stuck.

1. When 4.21 grams of hydroxide are added to 250 mL of water, the temperature rises by 4.14 °C. Assume that the density and specific heat of the dilute aqueous solutions are the same as those of H2O and calculate the molar heat of solution of potassium hydroxide.

**Would this eq be right?
q=(254.21g)(4.184J/gK)(4.14K)

2.Calculate the standard heat of formation of copper(I) oxide using the following data:
CuO → Cu + ½ O2 ΔH° = 157.3 kJ/mol
4 CuO → 2 Cu2 O + O2 ΔHo = 292.0 kJ/mol

Answer 1

1. Ignore the mass of the hydroxide in your calculation- it isn't going to make enough of a difference to the answer.
I assume, since you've given it to me, that 4.184 Jk^-1g^-1 is the specific heat capacity of water?
In that case, The enthalpy of solution is given by:
4.184 Jk^-1g^-1 x 250g x 4.14 k = 4330.44 J.

HOWEVER, this is per 4.21 g of KOH.
The molar mass of KOH is 56.11 gmol^-1.
How many moles is 4.21 g?
(4.21 g)/(56.11 gmol^-1) = 0.075 moles.

So, 4330.44 J per 4.21 g = 4330.44 J per 0.075 moles, = 4330.44/0.075 = 57739.2 Jmol^-1 = 57.7392 kJmol^-1.

2. You should draw a Hess cycle in order to better visualise this one. I can't do that on here, so we'll just have to struggle.

Firstly, make sure that you have the right number of reactants to make one mole of Cu2O:

4CuO -> 2Cu2O + O2: deltaH = 292.0 kJ/mol, becomes:
2CuO -> Cu2O + 1/2O2: deltaH = 292.0/2 = 146.0 kJ/mol.

Note that the equation now starts with 2 CuO
So, the other equation needs to END UP making 2 CuO:

CuO -> Cu + 1/2 O2: deltaH = 157.3 kJ/mol, becomes,
2Cu + O2 -> 2CuO: deltaH = - 314.6 kJ/mol.
NOTE: We have doubled the number of reactants, thus doubling deltaH, but we have also reversed the equation, so the positive deltaH value is now a negative value.

Now, we can combine these two equations, to give us the following:
2Cu + O2 -> 2CuO -----> 2CuO -> Cu2O + 1/2O2.
This is the reaction for which you were asked to calculate deltaH. So, you simply add the two deltaH values for the two parts that make up this equation:

(-314.6) + (146.0) = -168.6 kJ/mol. This is the enthalpy change per mole of Cu2O produced.

The important thing here is to make sure that the two parts that go together to make your final equation, all balance. Double check my workings, don't just copy them- I'm not infallible.

Answer 2

1)Not sure but based on the formula of q=mct, it sounds just about right.

2)i) Reverse first eq and multiply by 2
ii) Divide second eq by half
( Perform the same for the heat values and add them up to get the value of the standard heat of formation for Cu(1)oxide)

overall eq: 1/2O2 +2Cu -> Cu2O
Notice that the overall eq complies to its definition