Reactions with change in the number of moles. The decomposition of ammonia is co

Question

Reactions with change in the number of moles. The decomposition of ammonia is conducted in a chemistry lab, using the sealed glass apparatus shown in the picture. The reaction is described by the following stoichiometric equation NH_3 rightarrow 1/2 N_2 + 3/2 H_2 and is catalyzed by a metal filament. At the beginning of the experiment, the sealed flask contained 0.5 moles of NH_3 at a total pressure of 0.3 atm. After 30 min, the amount of NH_3 remaining in the flask dropped to 0.2 moles. Calculate the final pressure of the sealed flask. Assume that the gases can be considered ideal.

Explanation / Answer

1 mole NH3 ---> 1/2 N2 +3/2 H2

0.3 moles NH3----> (0.3 * 1/2) + (0.3 * 3/2) [AFTER REACTION OCCURRED]

since 0.3 moles of NH3 are consumed

0.2 moles NH3 remaining---> 0.15 moles N2 + 0.45 moles H2

So total pressure after the reaction has occurred would be due to [0.2 moles NH3 + 0.15 moles N2 + 0.45 moles H2 = ] 0.8 moles of all gases

Ideal gas equation PV = nRT

since flask is sealed, volume remains constant and even temperature is constant during the reaction and R is anyways Univeral gas constant

So pressure directkly depends upon "n" the number of moles

P1 = n1
P2 n2

0.3 = 0.5

x 0.8

x = 0.48 atm

So the final pressure would be 0.48 atm

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