Chemistry help pleaseee !!!! 🙏🙏🙏

Hi guys I’ve been sat trying to understand how to work this out for agesss but even looking at the mark scheme I don’t get it. I’ve attatched a photo but basically you have to write down the enthalpy change in KJmol-1 and then the enthalpy change of solution in KJmol-1. I get how to calculate the enthalpy change of solution but can’t for the life of me understand how to work the other one out. Any help would be SEVERLY appreciated !!! X

Reply 1

jjeeeeeea

12

I assume it is a born haber cycle? I don't see any attached image maybe you could post in comments?

Reply 2

Kiwiwatermelon

OP

2

Hi yes so sorry about that it’s not uploading . The information given in the question is:
Equation 22.1 H,SO (aq) + 2NaOH(aq) -› Na,SO (aq) + 2H,0(1)
This is a neutralisation reaction. A student carries out an experiment to determine the enthalpy change AH, and uses this value to deduce the enthalpy change of neutralisation. The student measures out two solutions: 25.0 cm3 of 1.60 moldm-3 H2SO4. 55.0 cm3 of 1.50 moldm-3 NāOH(aq) .The student mixes the two solutions. The temperature increases by 13.0°C.
1) Calculate the enthalpy change of in KJmol-1 ( this is the bit I can’t do )
2) Calculate the enthalpy change of neutralisation in KJmol-1 ( managed to do this but ).
Thank you :smile:

Reply 3

lukeb44

11

Original post by Kiwiwatermelon

Hi yes so sorry about that it’s not uploading . The information given in the question is:
Equation 22.1 H,SO (aq) + 2NaOH(aq) -› Na,SO (aq) + 2H,0(1)
This is a neutralisation reaction. A student carries out an experiment to determine the enthalpy change AH, and uses this value to deduce the enthalpy change of neutralisation. The student measures out two solutions: 25.0 cm3 of 1.60 moldm-3 H2SO4. 55.0 cm3 of 1.50 moldm-3 NāOH(aq) .The student mixes the two solutions. The temperature increases by 13.0°C.
1) Calculate the enthalpy change of in KJmol-1 ( this is the bit I can’t do )
2) Calculate the enthalpy change of neutralisation in KJmol-1 ( managed to do this but ).
Thank you :smile:

Ok so to work out the first enthalpy change you just do normal q = mcΔT, I assume you're fine with this. The enthalpy change of neutralisation is defined as the enthalpy change when 1 mol of water is formed from a neutralisation reaction - this reaction forms 2 mol of water. Meaning we need to divide by 2 to find the enthalpy change of neutralisation. Hope this helps :smile:

Chemistry help pleaseee !!!! 🙏🙏🙏

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