Calculate the internal energy at 298 K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is `-42.0 kJ mol^(-1)`.
(Give : `R = 8.314 JK^(-1) mol^(-1)`)

Explanation

Formation of ammonia :
`(1)/(2)N_(2(g))+(3)/(2)H_(2(g))to NH_(3(g))`
`therefore Delta n` = (No. of moles of gaseous product) - (No. of moles of gaseous reactants)
`therefore Delta n =1-[(1)/(2)+(3)/(2)]`
= -1
Formula, `W =-Delta nRT`
`=-(-1)xx8.314xx298=2477 J`
`W = 2.477 kJ " " (W = P Delta V)`
`Delta H=-42.0 kJ mol^(-1)`
`therefore Delta H = Delta U+ Delta nRT " " (Delta H = Delta U+P Delta V)`
`-42.0 = Delta U+ 2.477`
`Delta U=-42-2.77=-44.47 kJ`