6 replies to this topic

[dfx]

How do you work this out?

I've got to the bit where you do mcdeltaT using 0.04 Kg and 6.5 centigrade and 4.18 JK/KgK, but then what? Any help appreciate.d. cheers.

[Soph.E.]

I have no idea either so someone PLEASE reply quickly!!! The answer booklet doesn't tell you anything either!

[dfx]

Just managed to figure it out.

After doing mcdeltaT you get 1.0868 KJ released.

Calculate the no. of moles of acid using n = cv, and you get 0.02 (20/1000 x 1)

so 1.0868 KJ released by 0.02 mol
x released by 1 mol

And you cross multiply as usual and end up with -54 KJ/mol (-ve because its exothermic - resulting NaCl is higher temp. than the reactant solutions).

[Gavers]

I know this!!!

Heres what I got, from the start

E_h = C m t
= 4.18 (specific heat capacity of water) * 0.04 (mass of the two liquids) * 6.5 (Temp change in the liquids)

= 1.0868 kJ


0.02 moles give us 1.0868 kJ
1 mole gives us x kJ

Cross multiply to find the answer, so it is...


0.02x = 1 * 1.0868

x = 1.0868
0.02 (<-- That's supposed to be underneath the other thing)

x = 54.34 kJ/mol, and thats yer answer

Hope ya followed it okay, I actually made a daft mistake right at the start and got 4.5 as the temperature hange, d'oh!!

Gavers =)

[dfx]

Cheers m8.. thanks

[Gavers]

No probs, but ya beat me to it

I forgot about working out the concentration, I didn't even put that calculation down

[Soph.E.]

Aaahh so THAT'S how you do it... silly mistake really - always, always I do not look at the question properly! Thank you!