Suppose you mix 4.1 mol of a monatomic gas at 21.0 Solution The number of moles

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The number of moles of ammonia in equilibrium is: n(NH3) = m(NH3) / M(NH3) = 1.09g / 17g/mol = 6.41×10?² mol Using ideal gas law you can compute partial pressure for ammonia p(NH3) = n(NH3)·R·T/V = 6.41×10?² mol · 8.3145 Pam³mol?¹K?¹ · (300 + 273)K / 1.0×10?³m³ = 305470 Pa = 3.015 atm According to reaction equation and law of mass action partial pressures in equilibrium are related as: Kp = p(NH3) / ( p(N2)·p(H2)³ ) According to reaction equation the amount hydrogen formed due to reaction is three times as much as the amount of nitrogen. Because neither N2 nor H2 was in the flask initially, the equilibrium amounts are related as: n(H2) = 3·n(N2) As shown above the partial pressures are proportional to number of moles. Hence, p(H2) = 3·p(N2) Substitute p(H2) to equilibrium expression and solve for p(N2): Kp = p(NH3) / ( p(N2)·(3·p(N2))³ ) = p(NH3) / ( 81·p(N2)4 ) => p(N2) = v( p(NH3) / (81·Kp) ) = v( 3.015 / (81 · 4.34×10?²) ) = 1.711 atm => n(N2) = p(N2)·V / (R·T)/V = 1.711·101325 Pa ·1.0×10?³m³ / (8.3145 Pam³mol?¹K?¹ · (300 + 273)K ) = 3.64×10?³ mol => m(N2) = n(N2)·M(N2) = 3.64×10?³ mol · 28g/mol = 1.02 g 2) As shown in part 1) n(H2) = 3·n(N2) = 3 · 3.64×10?³ mol = 10.92×10?³ mol => m(H2) = n(H2)·M(H2) = 10.92×10?³ mol · 2 g/mol = 0.22 g 3) Conservation of mass requires, that the total mass of gas in the flask is constant. So the initially mass of ammonia is the same as the total mass of ammonia, nitrogen and hydrogen in equilibrium: m(NH3)0 = m(NH3) + m(N2) + m(H2) = 1.09g + 1.02g + 0.22g = 2.33 g 4) According to Dalton's law of partial pressure the total pressure equals the sum of partial pressures: p_total = p(NH3) + p(N2) + p(H2) = p(NH3) + p(N2) + 3·p(N2) = p(NH3) + 4·p(N2) = 3.015atm + 4·1.711atm = 9.859 atm

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