Calculate the work done on the system in Joules when 2.0 moles of `N_(2)` reactes with 6.0 moles `H_(2)`, to form `NH_(3)` against a pressure of 1 .0 atm at `27^(@)` C. `N_(2)(g)+3H_(2)(g)to2NH_(3)(g)`
(Given : R= `8 J//mol K`)

Calulate the work done (in cal.). When 1.0 mole of N_(2)H_(4) decomposes completely against a pressure of 1.0 atm at 27^(@)C (Given R=2 cal//mol//K) N_(2)H_(4)(l)toNH_(3)(g)+N_(2)(g)

Calculate the expression for K_(C) "and" K_(P) if initially a moles of N_(2) "and" B "moles of" H_(2) is taken for the following reaction. N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) (Deltanlt0) (P,T,V given)

2 mol of N_(2) is mixed with 6 mol of H_(2) in a closed vessel of 1L capacity. If 50% of N_(2) is converted into NH_(3) at equilibrium, the value of K_(C ) for the reaction N_(2)(g0+3H_(2)(g)hArr 2NH_(3)(g) is

In Haber's process 50.0 g of N_(2)(g) and 10.0 g of H_(2)(g) are mixed to produce NH_(3)(g) . What is the number of moles of NH_(3)(g) formed?

One mole of N_(2) (g) is mixed with 2 moles of H_(2)(g) in a 4 litre vessel If 50% of N_(2) (g) is converted to NH_(3) (g) by the following reaction : N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) What will the value of K_(c) for the following equilibrium ? NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)

One mole of N_(2) (g) is mixed with 2 moles of H_(2)(g) in a 4 litre vessel If 50% of N_(2) (g) is converted to NH_(3) (g) by the following reaction : N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) What will the value of K_(c) for the following equilibrium ? NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)