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 A ccording to the balanced equation `alpha` moles of `NH _(3)` decompose to produce `alpha //2` mol of `N _(2) ` and `3 alpha //2` mol of ` H _(2).`
`2NH _(3) hArr N _(2) + 3 H _(2)`
`{:("Initial moles", 1,0, 0),("Moles at equilibrium", 1-alpha, alpha//2, 3alpha//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 the same temperature T") = (P _(a))/(P _(b))`
`(1)/( 1 + alpha ) = (31 atm)/(50 atm) implies alpha = 0. 613,%` decompostition of ammonia `= 0. 613 xx 100 = 51. 3 %`