One mole of nitrogen is mixed with 3 mole of hydrogen in a closed 3 litre vessel. `20%` of nitrogen is converted into `NH_(3)`. Then `K_(C)` for the `(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)hArr NH_(3)` is

One mole of nitrogen is mixed with 3 mole of hydrogen in a closed 3 litre vessel 20 % of nitrogen is converted into NH_3 .Then what is the K_C for (1)/2(N_2) + (3)/2(H_2) ⇌ NH_3

1 mole of nitrogen is mixed with 3 moles of hydrogen in a sqrt3 litre container where 66.67% of nitrogen is converted into ammonia by the following reaction : N_2(g)+3H_2(g) hArr 2NH_3(g) ,Then the value of K_C for the reaction NH_3(g) hArr 1/2N_2(g) + 3/2H_2 will be :

One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) , then calculate the equilibrium constant, K_(c) in concentration, units. What will be the value of K_(c) for the following equilibrium? 1/2 N_(2)(g)+3/2H_(2)(g)hArrNH_(3)(g)

1 mol of nitrogen is mixed with 3 mol of hydrogen in a 4 L container. If 0.25% of nitrogen is converted to ammonia by the following reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) then calculate the equilibrium constant K_(c) in concentration units. What will be the value of K_(c) for the following equilibrium? 1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)

One mole of nitrogen is mixed with three moles of hydrogen in a 4 litre container. If 0.25 per cent of nitrogen is converted into ammonia by the following reaction N_2(g) +3H_2 hArr 2NH_3(g) calculate the equilibrium constant of the reaction in concentration units. What will be the value of K for the following reaction? (1)/(2) N_2 (g) + (3)/(2) H_2 hArr NH_3 (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)