When 40.0 ml of 0.200 m hcl at 21.5°c is added to 40.0 ml of 0.200 m naoh also at 21.5°c in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°c. assume that the volumes are additive, the specific heat of the solution is 4.18 jg -1°c -1 and that the density of the solution is 1.00 g ml -1 calculate the enthalpy change, δh in kj for the reaction:

The enthalpy change for the reaction is ΔH = - 54.3 kJ/mol

Explanation:

The reaction between HCl and NaOH is a neutralization reaction:

Heat released during neutralization = Heat gained by water

i.e.

where:

m = total mass of solution

ΔT = change in temperature = 22.8 - 21.5 = 1.3 C

c = specific heat = 4.18 J/g C

As per equation (1): qrxn = -434.7 J

The reaction enthalpy ΔH is the heat released per mole of acid (or base)

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.