When pure sulfuric acid is dissolved in water,heat is evolved.To a calorimeter containing 300g of water at 20degree C,10.65g of sulfuric acid also added at the same temperature.The change in temperature,which was monitored by a digital probe with negligible heat capacity ceased when it reaches 26.55 degree C.If specific heat of mixture is 4.184 J/g.K and small heat capacity of calorimeter ignored,what is the heat evolved/mole of sulfuric acid? I did this question and I got -75.7kJ/mole. I would like to know if my answer is correct. Thanks
3 answers
Looks good to me.
But in the answer options:
a) -27.4 kJ
b) -72.8 kJ
c) -78.4 kJ
d) -84.6kJ
I attempted this problem in the other way.When using the q =mcdT,I added the mass of both water and sulfuric acid.
q = (300g +10.65g)(4.184J/g.K)(6.55C)
= 8513.43 J
Then, I calculated the mole of sulfuric acid ,n = 10.65g/(98.08g/mol) = 0.1086mol
Since the question asks for heat evolved/mole of H2SO4,so
q sulfuric acid = 8513.43J/(0.1086mol)x 1 mol of H2SO4
= -78.4 kJ(option c)
Need detailed explanation.Thank you.
a) -27.4 kJ
b) -72.8 kJ
c) -78.4 kJ
d) -84.6kJ
I attempted this problem in the other way.When using the q =mcdT,I added the mass of both water and sulfuric acid.
q = (300g +10.65g)(4.184J/g.K)(6.55C)
= 8513.43 J
Then, I calculated the mole of sulfuric acid ,n = 10.65g/(98.08g/mol) = 0.1086mol
Since the question asks for heat evolved/mole of H2SO4,so
q sulfuric acid = 8513.43J/(0.1086mol)x 1 mol of H2SO4
= -78.4 kJ(option c)
Need detailed explanation.Thank you.
I also got -78.4kjmol^-1. Is this correct?
2 answers