Question

Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The ΔHf of Ca(OH)2 is -986.09 kJ/mol and liquid H2O is -285.8 kJ/mol.)

Ca(s) + 2H2O(l) Ca(OH)2 (s) + H2(g)

ΔHrxn =

The reaction is ?

Can anyone help cuz I suck at this.

I think it's -414.4kJ and endothermic reaction. I think.

Answer 1

we need to break this apart and start step by step.

I hope you take the time to read this as I've tried to explain this in the most straightforward way possible. I like straightforwardness lol.

the heat of reaction tells us how much heat was either released or absorbed by our reaction.

when i say it this way, I mean if we take the difference of energies for the products - the reactants. if we get a negative delta H that means we have released energy. this means exothermic reaction.
see below



the meaning of this is that we have lost energy in the form of heat and our products are at a lower energy than our reactants. this means they are more stable.





if we have positive delta h this means our reactants have a higher energy than our products. this means that we needed to put in energy to get our reaction to go that way.


Here is the formula:



\[\Delta H_{f} = \Delta H_{products}-\Delta H_{reactants} \]


and here is our reaction:

\[Ca(s) + 2H2O(l) \rightarrow Ca(OH)2 (s) + H2(g)\]

we can see that this is balanced.

There is another thing you need to know. the heat of formation Hf , means how much energy we would need to form that substance. if you notice, elements that occur naturally like H2 gas and Ca(s) would have an Hf of 0. why? because these elements occur naturally so there isn't any


Check this out

\[\Delta H_{f} = \frac{ kj }{ mol }\]

important thing here: kj/mol means kilojoules per mole of energy released. this means PER MOLE. so if we had say 0.5 moles of our substance we would have to multiply this by the number you're given. usually these are found on the table.

With these thing in mind this should help you solve the problem.