Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would effuse out of the same pinhole in 1.00 hour? Assume both gases are at the same temperature and pressure.

0.0019mol

Explanation:

To solve this problem, first we calculate for the rates of the effusion:

For CO2,

Number of mole = 0.00150 mol

Time = 1h

R1 = mol / time

R1 = 0.0015/1

R1 = 0.0015mol/h

Molar Mass of CO2 (M1) = 44.0 g/mol

For N2,

R2 =?

Molar Mass of N2 (M2) = 28.0 g/mol

Using Graham's law, we can obtain Rate (R2) of N2 as follows:

R1/R2 = √(M2/M1)

0.0015/R2 = √(28/44)

0.0015/R2 = 0.7977

R2 = 0.0015 /0.7977

R2 = 0.0019mol/h

But rate = mole /time

Time = 1h

Rate od N2 = 0.0019mol/h

0.0019 = mole /1

Mole = 0.0019mol

Therefore, the number of mole of N2 is 0.0019mol

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.