97 GAS DIFFUSION AND EFFUSIONA flammable gas made up only of carbon and hydrogen is found to effuse through a porous barrier in 1.50 minutes. Under the same conditions of temperature and pressure it takes an equal volume of bromine vapor 4.73 minutes to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest that this gas might be.

98 GAS DIFFUSION AND EFFUSION

99 GAS DIFFUSION AND EFFUSION

100 GAS DIFFUSION AND EFFUSIONGas X has a molar mass of 72 g/mol and Gas Y has a molar mass of 2 g/mol. How much faster or slower does Gas Y effuse from a small opening than Gas X at the same temperature? Answer: next slide

101 GAS DIFFUSION AND EFFUSIONGas X has a molar mass of 72 g/mol and Gas Y has a molar mass of 2 g/mol. How much faster or slower does Gas Y effuse from a small opening than Gas X at the same temperature? Answer: Gas Y will effuse six times faster than the heavier Gas X.

102 GAS DIFFUSION AND EFFUSIONGas X has a molar mass of 72 g/mol and Gas Y has a molar mass of 2 g/mol. How much faster or slower does Gas Y effuse from a small opening than Gas X at the same temperature? Solution: Graham's Law can be expressed as rX(MMX)1/2= rY(MMY)1/2where rX= rate of effusion/diffusion of Gas X MMX= molar mass of Gas X rY= rate of effusion/diffusion of Gas Y MMY= molar mass of Gas Y We want to know how much faster or slower Gas Y effuses compared to Gas X. To get this value, we need the ratio of the rates of Gas Y to Gas X. Solve the equation for rY/rX. rY/rX= (MMX)1/2/(MMY)1/2rY/rX= [(MMX)/(MMY)]1/2Use the given values for molar masses. rY/rX= [(72 g/mol)/(2)]1/2rY/rX= [36]1/2rY/rX= 6

103 Using PV = nRTHow much N2is req’dto fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? R = 0.082057 L•atm/K•molSolution 1. Get all data into proper units V = 27,000 L T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm

104 Using PV = nRTHow much N2is req’d to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC? R = 0.082057 L•atm/K•molSolution 2. Now calc. n = PV / RT n = (0.98 atm)(2.7 x 104L)(0.0821 L•atm/K•mol)(298 K)n = 1.1 x 103mol (or about 30 kg of gas)

105 Gases and Stoichiometry2 H2O2(liq) 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2in a flask with a volume of 2.50 L. What is the pressure of O2at 25 oC? Of H2O? Solution Strategy: Calculate moles of H2O2and then moles of O2and H2O. Finally, calc. P from n, R, T, and V.

106 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2in a flask with a volume of 2.50 L. What is the pressure of O2at 25 oC? Of H2O? Solution 1.1 g H2O2•1 mol34.0 g0.032 mol0.032 mol H2O2•1 mol O22 mol H2O2= 0.016 mol O2

107 Gases and Stoichiometry 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Decompose 1.1 g of H2O2in a flask with a volume of 2.50 L. What is the pressure of O2at 25 oC? Of H2O?