A 100 mL sample of .300 M NaOH is mixed with a 100 mL sample of .300 M HNO3 in a

Question

A 100 mL sample of .300 M NaOH is mixed with a 100 mL sample of .300 M HNO3 in a coffee cup calorimeter. If both solutinos were initially at 35 degrees celcius and the temperature of the resulting solution was recorded at 37 degrees Celcius, determine ?H?rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HNO3. (assume--no heat is lost to the calorimeter or the surroundings and the densith and heat capacity of the resulting soltion are the same as water). The answer is -55.7kJ/mol NaOH. I just need an explanation of the steps that will get me to this answer so that I can review it for the exam.

Explanation / Answer

The equation for the neutralisation reaction is: NaOH + HNO3 --> NaNO3 + H2O Therefore in the reaction the final amount of NaNO3 by doing volume times concentration 0.1 x 0.3 = 0.03moles. You can then find out the mass of the NaNO3 which is moles x Molecular mass: (23+14+16+16+16) x 0.03=2.55g The amount of water that was in the coffee cup which had an increase in temperature is 100ml+100ml = 200ml which is 0.2dm3 if you use the forumular to find the energy taken to increase the temperature: q = mct c = specific heat capacity of water which is 4.2 t = change in temp q = 2.55 x 4.2 x 2= 21.42kJ Now the qs asks you to find delta H which is the amount of energy needed for the neutralisation of one mole! we just calculated the amount for 0.03 moles. do what you have to do it: 21.42/0.03 = 714kJ As it was an exothermic reaction the answer is -714kL

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