heat of neutralization?

Question

when sodium hydroxide solution and hydrochloric acid are mixed together,a rise in temperature is noted.the volume of each solution is 100cm^3 and their concentration is 1.0mol/dm^3.the heat produced is calculated as 5.7 kJ.

(a)calculate the heat of neutralization

(b)if the hydrochloric acid is replaced by nitric acid,predict the value of heat of neutralization.explain why?

any explanations are appreciated

thanks guys

Simon M · Accepted Answer

The concentraion is 1 mol per litre, so in 100 cc there will be 0.1 mol.  Therefore, you get 5.7 kJ evolved from 0.1 mol, so the heat of neutralisation will be 57 kJ/mol.

The standard enthalpy of neutralisation is given by E = cmd where E is the enthalpy, c is the specific heat of water, and d is the temperature delta, and m is the mass of solution.  Therefore changing from hydrochloric to nitric acid will not make any difference to the enthalpy of neutralisation.  Both are strong acids.

If you changed from hydrochloric to acetic, which is a weak acid and only partially ionised in water....then it would be slightly more complex.

? · Answer

the warmth of neutralization is the warmth of reaction whilst one gram equivalent of acid reacts with one gram equivalent of base to create one mole of salt and one mole of water. be conscious : this is one gram equivalent and not one mole. answer : 5.80 two kilojoules quantity of hcl = mass/density = a hundred/a million = a hundred style of moles of hcl = a million.02 x a hundred/one thousand = 0.102 one mole of hcl whilst neuteralised by utilising one mole of NaoH evolves warmth = fifty seven.a million KJ So 0.102 mole of hcl is neturalised by utilising 0.102 moles of NaoH evolves warmth = fifty seven.a million x 0.102 = 5.80 two KJ

heat of neutralization?

2 Answers