Ammonia under a pressure of 15 atm at `27^(@)C` is heated to `347^(@)C` in a closed vessel in the presence of catalyst. Under the conditions, `NH_(3)` is partially decomposed according to the equation`2NH _(3) hArr N _(2) + 3H _(2).` The vessel is such that the volume remains effectively constant, whereas pressure increases at 50 atm. Calculate the percentage of `NH_(3)` actually decomposed.

Options

A. `61.3%`
B. `63.5%`
C. `65.3%`
D. `66.6%`

Answer

A

Explanation

Let `alpha`- be the degree of dissociation of ammonia under these conditions. According to the balanced equaiton `alpha` moles of `NH_(3)` decompose to produce `alpha//2 mol` of `N_(2)` mol and `3 alpha//2` mol of `H_(2)`.
`{:(,2NH_(3),hArr,N_(2),+,3H_(2)),("Initial moles",1,,0,,0),("Moles at equilibrium",1-alpha,,alpha//2,,3 alpha//2):}`
Total number of moles after decomposition `= 1 + alpha`. Using Gay Lussac.s law
`(P_(1))/(T_(1)) = (P_(2))/(T_(2)), P_(2) = P_(1) xx (T_(2))/(T_(1)) = 15 xx (273 = 347)(273 + 27) = 15 xx (620)/(300) = 31` atm
`("Number of moles without decomposition at temperature T")/("Number of moles after decomposition at the same temperature T") = (P_(a))/(P_(b))`
`(1)/(1+alpha) = ("31 atm")/("50 atm") rArr alpha = 0.613`, % decomposition of ammonia `= 0.613 xx 100 = 61.3%`