Calculating the standard enthalpy change for the following reaction:

have you drawn a hess cycle?

Hi, I'm trying to solve this question for chemistry, but have been going round in circles and can't get my head around it. I would be super grateful if anyone could help.2Li(s) 2H2O(l) → 2Li (aq) 2OH−(aq) H2(g)Calculate the standard enthalpy change for this reaction (above) using the following data: ∆Hf [LiOH(s)] = −487kJmol−1∆Hf [H2O(l)] = −286kJmol−1∆Hf [Li(s)] = 0kJmol−1LiOH(s) + aq → Li+(aq) + OH−(aq) ∆H = − 21kJmol−1

You just need to construct the desired equation from the other equations. It's a case of using the four rules of number, +-*/.

You need:

2Li(s) + 2H2O(l) → 2Li+(aq) + 2OH−(aq) + H2(g)

So you look for an equation that has Li(s) in it.

Although Li(s) is the element in its standard state and so no energy change is involved in making it, so move on.

The next species is water, H2O and you need 2 of them. he enthalpy of formation of water is given. BUT you are not making water (formation) you are destroying it (negative formation), so you double the negative value.

You now have:

2Li(s) + 2H2O → separate elements ∆H = 2 * -(-286) = +572kJ

Now add in the products: You need to make 2LiOH(s) and then dissolve the ions in water (this is two steps)

2Li(s) + 2H2O → separate elements → 2Li+(aq) + 2OH-(aq) ............. ∆Hf = 2*(-487) + 2*(-21) = -974 - 42 = -1016 kJ

Add this to +572 kJ and you get 572 - 1016 = -444kJ

You can forget about the hydrogen because it is an element in standard state.

Thank you, this makes sense. Can you please tell me if/how I could represent this in a Hess cycle?

You just need to construct the desired equation from the other equations. It's a case of using the four rules of number, +-*/.

You need:

2Li(s) + 2H2O(l) → 2Li+(aq) + 2OH−(aq) + H2(g)

So you look for an equation that has Li(s) in it.

Although Li(s) is the element in its standard state and so no energy change is involved in making it, so move on.

The next species is water, H2O and you need 2 of them. he enthalpy of formation of water is given. BUT you are not making water (formation) you are destroying it (negative formation), so you double the negative value.

You now have:

2Li(s) + 2H2O → separate elements ∆H = 2 * -(-286) = +572kJ

Now add in the products: You need to make 2LiOH(s) and then dissolve the ions in water (this is two steps)

2Li(s) + 2H2O → separate elements → 2Li+(aq) + 2OH-(aq) ............. ∆Hf = 2*(-487) + 2*(-21) = -974 - 42 = -1016 kJ

Add this to +572 kJ and you get 572 - 1016 = -444kJ

You can forget about the hydrogen because it is an element in standard state.