In the process of underwater welding, a gas bubble is intentionally formed around the weld to protect it from water until it has cooled. To do so, a mixture of gas is used containing mole fractions of χH2=0.70 (70% hydrogen), χCO2=0.25 (25% carbon dioxide), and χCO=0.05 (5% carbon monoxide). If the total pressure on this gas bubble is 2.6 atm, what are the partial pressures of the three component gases (in atm)?

See explanation

Explanation:

Total number of moles of gases in the mixture; 0.70 + 0.25 + 0.05 = 1 mole

Partial pressure= mole fraction of gas × total pressure

Mole fraction of hydrogen = 0.7/1 × 2.6 = 1.82 atm

Mole fraction of CO2 = 0.25/1 × 2.6 = 0.65 atm

Mole fraction of CO= 0.05/1 × 2.6 = 0.13 atm

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

i believe the answer is true.