A sample of gas contains 0.1200 mol of H2(g) and 0.1200 mol of O2(g) and occupies a volume of 11.5 L. The following reaction takes place:
H2(g) + O2(g)>H2O2(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
L

5.75L is the volume of the sample after the reaction

Explanation:

Based on the reaction, 1 mole of H2 reacts with 1 mole of O2 to produce 1 mole of H2O2.

As in the reaction, 0.1200 moles of H2 and 0.1200 moles of O2 are added, 0.1200 moles of H2O2 are produced.

Before the reaction, the moles of gas are 0.2400 moles and after the reaction the moles are 0.1200 moles of gas.

Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:

V1/n1 = V2/n2

Where V is volume and n are moles of 1, initial state and 2, final state.

Replacing:

V1 = 11.5L

n1 = 0.2400 moles

V2 = ?

n2 = 0.1200 moles

11.5L*0.1200 moles / 0.2400 moles = V2

V2 = 5.75L is the volume of the sample after the reaction

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 :