40% of a mixture of 0.2 mol of `N_2 ` and 0.6 mol of `H_2` react to give `NH_3` according to the eqution, `N_2(g)+3H_2(g) hArr 2NH_3(g)`, at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

40% of a mixture of 0.2 mol of N_(2) and 0.6 mol of H_(2) reacts to give NH_(3) according to the equation: N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

40% of a mixture of 2.0 mol of N_(2) and 0.6 mol of H_(2) reacts to give NH_(3) according to the equation: N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

40% of a mixture of 0.2 mole of N_2 and o.6 mole of H_2 react to give NH_3 according to the equation: N_ 2 ( g ) + 3 H_ 2 ( g ) ⇔ 2 N H_ 3 (g) at constant temperature and pressure. Then what is the ratio of the final volume to the initial volume of gases ?

For the reaction, N_2[g] + 3H_2[g] hArr 2NH_3[g] , Delta H = …

In the reaction, N_2(g) +3H_2(g) hArr 2NH_3(g) the value of the equilibrium constant depends on

If Ar is added to the equilibrium N_2(g) +3H_2(g) hArr 2NH_3 at constant volume , the equilibrium will