1. Under certain atmospheric conditions, the temperature of the surrounding air rises as a snowfall begins, because energy is released to the stmosphere as water changes to snow. What is the enthalpy change delta H for the freezing of 1000kg of water at 0.0 degree C to 1000kg of snow at 0.0 degree C?

Calculations:
delta H(fusion)=6.03kJ/mol
m=1000000g
n=1000000g/18.02g=55.493896mol
delta H=(delta H(fusion))(n)=(6.03)(55493.896)
delta H=334628.19kJ=3.35x10^5kJ

But according to my text book, "enthalpy changes for exothermic rxn are given a negative sign and enthalpy changes for endotheermic rxn are given a positive sign..." The trouble is the textbook has a positive answer(maybe wrong...)

So should the answer be negative, ie -3.35x10^5kJ?
If so, which part in the calculation did I do wrong? (I get a positive)

Molar enthalpy of fusion of H2O=6.03
would freezing make this value negative? But my teacher says this is always positive, no matter melting or freezing...I am confused...

Please help me! Thank you!

The heat of fusion is always positive since fusion is melting and melting is endothermic.
The opposite of fusion is freezing, so a negative sign can be placed in front of it to indicate exothermic release of heat for the enthalpy change of water. That would be technically legitimate.
The book leaves it as positive probably because they are just interested in the magnitude of the value.
Or, it may be positive to indicate the enthalpy change of the surrounding atmosphere which is opposite to the enthalpy change of the freezing water.

So is it acceptable to make delta H(freezing)=-6.03kJ/mol and do it this way?
Calculations:
delta H(freezing)=-6.03kJ/mol
m=1000000g
n=1000000g/18.02g=55.493896mol
delta H=(delta H(freezing))(n)=(-6.03)(55493.896)
delta H=-334628.19kJ=-3.35x10^5kJ

oic...thank you, Demotivator

2.A 9.18g sample of C2H5OH is burned in a bomb calorimeter. The temp. of the calorimeter rose from 21.30 to 32.54C. If the heat capacity of the calorimeter was 4.56kJ/C, what is the enthalpy change when 3.56 mol of C2H5OH is burned? (ANSWER: 916kJ)

Calculations:
How can I calculate delta H when 3.56 mol C2H5OH is burned?
I tried this way:
Surroundings:
Energy of calorimeter is absorbed:
q=c(delta T)=(4.56kJ/C)(11.24C)=51.2544kJ

System:
Energy of C2H5OH released=51.2544kJ
Therefore, enthalpy change= -51.2544kJ
But the answer is 916kJ...How can I calculate that? (and how come the answer is positive? exothermic=>negative...right?)

thank you!

2) 51.2544kJ  is the enthalpy per  9.18g sample. The question is for 3.56 moles so  convert to per mole, etc.

correct, exothermic is negative. But apparently the book is only interested in the magnitude and doesn't care whether exothermic or endothermic.

3)  If H is positive 104kJ it is endothermic and would go with reactants:
 104kJ + 3XX +4YY ->1244ZZ

If H is negative, then energy is a product:
3XX +4YY ->1244ZZ + 104kJ

The coefficients make a difference if the energy is just in units of kJ because energy depends on moles of  material.  If energy is expressed in kJ per mole, then that number is independent of the coefficients.

Let me put it this way.
2A -> B  H = 10kj or H = 5 kj /moleA
A  ->  1/2 B  H = 5kj   or H = 5 kj/moleA

therefore as a ratio it doesn't depend on the coefficients!

1) Adding the contributions of the individual heat capacities gives the total heat for the reaction.

2) The heat of reaction thus obtained is not the heat of neutralization (the acid and base do not react completely). You need to determine the limiting reactant and divide by the limiting moles to obtain the molar heat of neutralization.

I don't quite understand the equation
delta H = +/- |q(surroundings)|
What does it mean and why is it so complicated? (absolute value/ +/-...)

I thought that delta H= -q

does anyone understand this equation? thanks!

deltaH(system) = -q(of the surroundings)
q could be positive or negative. So, if q(surroundings) is negative, deltaH(system) is positive and vice versa.

delta H = +/- |q(surroundings)|
This is just a fancy and bad way of saying the same thing.
It means deltaH of the system has the same MAGNITUDE (thus the absoute sign) as the heat change of the surroundings,  but the sign of deltaH is either positive (endo) or negative (exothermic).
The +/- is there because the absolute sign destroys the internal sign of q(surroundings), so the "lost" sign has to be reintroduced as + or - depending on whether the system is exo or endo.

oh...fancy and bad way...ha
I also think that delta H=-q makes more sense! Thank you Demotivator!