Mixture of 0.1mol of NO,0.5mol of H2 and 0.1 mol of water is placed in a 1L vessel at 300K , at equilibrium the concentration of No is 0.06M. Calculate the concentration of N, H, and water.the equation is 2NO + 2H2 gives 2No+2H2O
Initial concentration:
(NO) = 0.1mol/1L = 0.1 M
(H2) = 0.5 mol/1L = 0.5 M
(H2O) = 0.1 mol/1L = 0.1 M
I don't believe the equation is correct. Please check your post. What is No?
To calculate the concentration of N, H, and water at equilibrium, we can use the stoichiometry of the given reaction.
First, let's determine the initial concentrations of NO, H2, and water:
- Initial concentration of NO = moles of NO / volume of the vessel = 0.1 mol / 1 L = 0.1 M
- Initial concentration of H2 = moles of H2 / volume of the vessel = 0.5 mol / 1 L = 0.5 M
- Initial concentration of water = moles of water / volume of the vessel = 0.1 mol / 1 L = 0.1 M
Now, let's determine the change in concentrations of NO, H2, and water at equilibrium using the given balanced equation:
2NO + 2H2 → 2NO + 2H2O
Here, the stoichiometric coefficient of NO is 2, indicating that for every 2 moles of NO, we obtain 2 moles of NO (as a product). This means that the concentration of NO will remain the same at equilibrium since the number of moles of NO remains constant.
Therefore, the concentration of NO at equilibrium remains 0.06 M.
Next, let's consider the stoichiometric coefficient of H2, which is also 2. This means that for every 2 moles of H2, we obtain 2 moles of H2O (as a product).
Since the stoichiometric coefficient of H2 matches that of H2O, the concentration of H2 will also be 0.06 M at equilibrium.
Finally, let's consider the stoichiometric coefficient of water (H2O), which is also 2. However, initially, we have 0.1 mol of water, and each mole of water forms 2 moles of H2O in the reaction. Therefore, the concentration of water at equilibrium will be:
Concentration of water = (initial moles of water - change in moles of water) / volume of the vessel
= (0.1 mol - 0.1 mol) / 1 L
= 0 M
Hence, the concentration of water at equilibrium is 0 M.
To summarize the results:
- Concentration of NO at equilibrium = 0.06 M (remains the same)
- Concentration of H2 at equilibrium = 0.06 M (remains the same)
- Concentration of water at equilibrium = 0 M