The enthalpy of fusion of methanol (ch3oh) is 3.16 kj/mol. how much heat would be absorbed or released upon freezing 25.6 grams of methanol?

5.04 kJ

Explanation:

We are given the enthalpy of fusion of methanol as 3.16 kJ per mol of methanol. Thus, what we need is to determine is the number of mol in 25.4 g of methanol and multiply by this enthalpy.

mol methanol = mass / MW CH₃OH = 25.6 mol / 16.04 g/mol = 1.60 mol

ΔH = 1.60 mol x 3.16 kJ/mol = 5.04 kJ

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 :