Graham’s Law of Diffusion

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Presentation transcript:

Graham’s Law of Diffusion

Graham’s Law of Diffusion HCl NH3 NH4Cl(s) 100 cm 100 cm Choice 1: Both gases move at the same speed and meet in the middle.

Diffusion NH4Cl(s) HCl NH3 81.1 cm 118.9 cm Choice 2: Lighter gas moves faster; meet closer to heavier gas.

Graham’s Law Diffusion Effusion Spreading of gas molecules throughout a container until evenly distributed. Effusion Passing of gas molecules through a tiny opening in a container Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

KE = ½mv2 Graham’s Law Speed of diffusion/effusion Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly. Larger m  smaller v KE = ½mv2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Graham’s Law Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas A’s speed to Gas B’s speed. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Diffusion

Gas Diffusion and Effusion Graham's law governs effusion and diffusion of gas molecules. Rate of effusion is inversely proportional to its molar mass. Thomas Graham (1805 - 1869)

Diffusion and Effusion To use Graham’s Law, both gases must be at same temperature. diffusion: particle movement from high to low concentration NET MOVEMENT effusion: diffusion of gas particles through an opening For gases, rates of diffusion & effusion obey Graham’s law: more massive = slow; less massive = fast

Diffusion Particles in regions of high concentration spread out into regions of low concentration, filling the space available to them.

Weather and Diffusion LOW Air Pressure HIGH Air Pressure Map showing tornado risk in the U.S. Highest High

Graham’s Law Calculation Rate 1 = √M 2 Rate 2 √M 1 Where Rate1 = rate of effusion of the 1st gas Rate2 = rate of effusion of the 2nd gas M1 = molar mass of the 1st gas M2 = molar mass of the 2nd gas

Graham’s Law Practice Problems If neon gas travels at 400 m/s at a given temperature, calculate the velocity of butane, C4H10, at the same temperature. Nitrogen gas effuses through an opening 1.59 times faster than does an unknown gas. a. Calculate the molecular mass of the unknown gas. b. Make a reasonable prediction as to what the unknown gas is.