Exercise 5.53 with feedback You may want to reference(pages 201-205) Section 5.8

Question

Exercise 5.53 with feedback You may want to reference(pages 201-205) Section 5.8 while completing this problem Part A Calculate the root mean square velocity of F2,Clh, and Br2 at 302 K. Express the velocities in meters per second to three significant figures separated by commas. 0? m/s Submit My Answers Give Up Part B Calculate the kinetic energy of F2, Cl2, and Br2 at 302 K Express the energies in joules to three significant figures separated by commas Submit My Answers Give Up Part C Rank the three halogens with respect to their rate of effusion. Rank from fastest to slowest. To rank items as equivalent, overlap them. RosotHolp Fastest Slowest Submit My Answers Give Up

Explanation / Answer

rms velocity(Vrms) of F2 = sqrt(3RT/M)

R = 8.314 j.k-1.mol-1

T = 174 C = 302 k

M = molar mass of F2 = 38 g/mol

Vrms = sqrt(3*8.314*302/(38*10^-3))

      = 445.23 m/s

rms velocity(Vrms) of Cl2 = sqrt(3RT/M)

R = 8.314 j.k-1.mol-1

T = 174 C = 302 k

M = molar mass of Cl2 = 71 g/mol

Vrms = sqrt(3*8.314*302/(71*10^-3))

      = 325.7 m/s

rms velocity(Vrms) of Br2 = sqrt(3RT/M)

R = 8.314 j.k-1.mol-1

T = 174 C = 302 k

M = molar mass of Br2 = 160 g/mol

Vrms = sqrt(3*8.314*302/(160*10^-3))

      = 216.97 m/s

kinetic energy of F2 = 1/2mv^2

m = mass of F2 = 38 g/mol

v = vrms = 445.23 m/s

   = 1/2*(38*10^-3)*445.23^2

   = 3.77*10^3 joule

kinetic energy of Cl2 = 1/2mv^2

m = mass of Cl2 = 71 g/mol

v = vrms = 325.7 m/s

   = 1/2*(71*10^-3)*325.7^2

   = 3.77*10^3 joule

kinetic energy of Br2 = 1/2mv^2

m = mass of Br2 = 160 g/mol

v = vrms = 325.7 m/s

   = 1/2*(160*10^-3)*216.97^2

   = 3.77*10^3 joule

part C

rate of effusion of rate is inversily proportional to molarmass of the molecule.

Br2 = 160 g/mol

Cl2 = 71 g/mol

F2 = 38 g/mol

so that,

Fastet

   F2 > Cl2 > Br2

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