`NH_3` is heated initially at 15 atm from `27^@C` to `127^@C` at constant volume.At `127^@C` equilibrium is established.The new pressure at equilibrium at `127^@C` becomes 30 atm for the reaction `2NH_3(g)hArrN_2(g)+3H_2(g)` Then find the % of moles of `NH_3` actually decomposed.

`NH_3` is heated initially at 15 atm from `27^@C` to `127^@C` at const

1.	`NH_3` is heated initially at 15 atm from `27^@C` to `127^@C` at constant volume.At `127^@C` equilibrium is established.The new pressure at equilibrium at `127^@C` becomes 30 atm for the reaction `2NH_3(g)hArrN_2(g)+3H_2(g)` Then find the % of moles of `NH_3` actually decomposed.
Answer» Correct Answer - 50 `{:(2NH_3hArr,N_2+,3H_2),(a,0,0),(a-2x,x,3x):}` `P_1/T_1=P_2/T_2implies15/300=P_2/400implies P_2=20` atm Now `(a+2x)/a=30/20` 2a+4x=3a `implies x=1/4a " " :.` % of `NH_3` decomposed =`(2x)/axx100=50%`

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`NH_3` is heated initially at 15 atm from `27^@C` to `127^@C` at constant volume.At `127^@C` equilibrium is established.The new pressure at equilibrium at `127^@C` becomes 30 atm for the reaction `2NH_3(g)hArrN_2(g)+3H_2(g)` Then find the % of moles of `NH_3` actually decomposed.

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