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yrelhan4
SO3(g) decomposes according to the equation 2SO3(g) → 2SO2(g) + O2(g) A sealed container contains 0.50 mol of SO3 gas at 100°C and 2 atmospheres pressure. What would be the pressure in the container if the SO3 gas is decomposed completely according to the above equation and the temperature were maintained at 100°C ?
Because its a sealed container its volume is constant + and temperature is also constant: PV = nRT => V/RT = contstant too.. 2SO3 => 2SO2 + O2 0.5 mole 0.5 mole 0.25 mole V/RT = constant = n(initial)/P(initial) = n(final)/P(final) n(inital) = n(SO3) = 0.5 mole, P(initial) = 2atm, n(final) = n(SO2) + n(O2) => 0.5 mole + 0.25 mole = 0.75 mole, P(final) = x atm n(inital)/P(initial) = 0.5 mole/2 atm = 0.75 mole / x atm => 0.5x = 1.5 => x = 3 atm = P(final)