Question

A mixture of 0.100 Mol of NO, 0.0500 mol of H2, and 0.100 mol of H2O is placed in a 1.00L vessel. the following equilibrium is established: 2NO (g) + 2 H2 (g) -> N2(g) + 2 H2O (g) At equilibrium [NO] = 0.0620 M. calculate the equilibrium concentration of H2

Answer 1

To calculate the equilibrium concentration of H2 (hydrogen gas) at the given equilibrium condition, we can use the stoichiometry of the balanced chemical equation and the initial moles of NO, H2, and H2O, along with the change in the number of moles for each substance.

Given data:
Initial moles of NO (n(NO)) = 0.100 mol
Initial moles of H2 (n(H2)) = 0.0500 mol
Initial moles of H2O (n(H2O)) = 0.100 mol
Equilibrium concentration of NO ([NO]) = 0.0620 M

The balanced chemical equation is: 2NO (g) + 2H2 (g) -> N2 (g) + 2H2O (g)

At equilibrium, the change in the number of moles of NO is (0.100 - 0.0620) mol = 0.0380 mol.

Since the stoichiometry of the equation shows that for every 2 moles of NO consumed, 2 moles of H2 are also consumed and produce 1 mole of N2 and 2 moles of H2O, the change in the number of moles of H2 is also 0.0380 mol.

Therefore, the moles of H2 at equilibrium are:
n(H2) = 0.0500 mol - 0.0380 mol = 0.0120 mol

Now, we can calculate the equilibrium concentration of H2 (Molar concentration):

Equilibrium concentration of H2 ([H2]) = moles of H2 (n(H2)) / volume of the vessel (V)
[H2] = 0.0120 mol / 1.00 L
[H2] = 0.0120 M

The equilibrium concentration of H2 is 0.0120 M.


Hope this helps.
willkiddhill