Consider the decomposition equilibrium for dinitrogen pentoxide: At a certain te
ID: 972836 • Letter: C
Question
Consider the decomposition equilibrium for dinitrogen pentoxide: At a certain temperature and a total pressure of 1.00 atm, the N2O5 is 0.50% decomposed (by moles) at equilibrium. (a). If the volume is increased by a factor of 10.0, will the mole percent of N2O5 decomposed at equilibrium be greater than, less than, or equal to 0.50%. Explain. (b) Calculate the mole percent of N2O5 that will be decomposed at equilibrium if the volume is increased by a factor of 10.0
2N2O5(g) ---> 4NO2 (g) + O2 (g)
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Explanation / Answer
a) Here, the forward reaction occurs with increase in number of moles. (we have 2 moles of reactant and 5 moles of product). If the volume of the gaseous reaction is increased , the pressure exerted by the molecules will decrease. Thus, the effect of increase of volume is equivalent to the effect of decrease of pressure.
So, if pressure is decreased (or the volume is increased) , then according to the Le Chatlier's principle,the equilibrium will shift in a direction in which the pressure increases or the no. of moles increases. So, the equilibrium will shift in the forward direction which is accompanied by increase in total number of moles.
Hence, rxn will proceed in forward direction and amount of N2O5 will decrease(i.e more amount of N2O5 will react to form product) or mole % of N2O5 decomposed will increase.
So, if the volume is increased by a factor of 10.0, the mole percent of N2O5 decomposed at equilibrium will be greater than 0.50%
b) Let initial volume = V
Moles of N2O5 reacted = 0.50
molarity of N2O5 reacted = moles / volume = (0.50 / V )
new volume of the system = 10V
moles of N2O5 reacted = molarity * volume = (0.5/V) * 10V = 5 moles
mol % of N2O5 that has decomposed at equilibrium if the volume is increased by a factor of 10.0 = 5%