40% of a mixture of 0.2 mole of N 2 and 0.6 mole of H 2 react to give NH 3 according to the equation N 2 ( g ) + 3 H 2 ( g ) ⇌ 2 NH 3 ( g ) at given temperature and pressure. Then the rati

Question

40% of a mixture of 0.2 mole of N 2 ​ and 0.6 mole of H 2 ​ react to give NH 3 ​ according to the equation N 2 ​ ( g ) + 3 H 2 ​ ( g ) ⇌ 2 NH 3 ​ ( g ) at given temperature and pressure. Then the ratio of the final volume to the initial volumes of gases are as

TestHub · Accepted Answer

N 2 ​ ( g ) + 3 H 2 ​ ( g ) ⇌ 2 N H 3 ​ ( g ) ( 0.2 − x ) ( 0.6 − 3 x ) 2 x Hence, 0.8 4 x ​ = 0.4 ⟹ x = 0.08 . Total number of moles at equilibrium = ( 0.2 − x ) + ( 0.6 − 3 x ) + 2 x = 0.8 − 2 x 0.8 − 2 × 0.08 = 0.8 − 0.16 = 0.64 Therefore, n i ​ n f ​ ​ = V i ​ V f ​ ​ = 0.8 0.64 ​ = 10 8 ​ = 5 4 ​ = 0.8 .

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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