One "mole" of `N_(2)` is mixed with three moles of `H_(2)` in a `4L` vessel. If `0.25% N_(2)` is coverted into `NH_(3)` by the reaction
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`, calculate `K_(c)`. Also report `K_(c)` for
`1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)`

For the reaction N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?

For the reaction N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?

For a reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))

For the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the units of K_(p) are …………

For the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the units of K_(p) are …………

(K_(c))/(K_(p)) for the reaction, N_(2)(g)+3H_(2)(g)hArr2NH_(3(g)) is

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)