-and-0.5-mol-o2(g)-at-25-%E2%88%98c-or-1-mol-h2o(g)-at-25-%E2%88%98c.jpeg "1 Mol H2(g) And 0.5 Mol O2(g) At 25 ∘c Or 1 Mol H2o(g) At 25 ∘c")
Reaction Equilibrium of 1 mol H2(g) and 0.5 mol O2(g) at 25°C
In this article, we will discuss the reaction equilibrium of 1 mole of hydrogen gas (H2) and 0.5 mole of oxygen gas (O2) at a temperature of 25°C. We will also compare it with the reaction equilibrium of 1 mole of water vapor (H2O) at the same temperature.
The Reaction
The reaction between hydrogen gas and oxygen gas is a combustion reaction, which produces water vapor:
2H2(g) + O2(g) ⇌ 2H2O(g)
Initial Conditions
Initially, we have 1 mole of hydrogen gas (H2) and 0.5 mole of oxygen gas (O2) at a temperature of 25°C.
Reaction Equilibrium
At equilibrium, the reaction will reach a state where the rates of forward and reverse reactions are equal. The equilibrium constant (Kc) for this reaction is:
Kc = [H2O]² / [H2]²[O2]
where [H2O], [H2], and [O2] are the concentrations of water vapor, hydrogen gas, and oxygen gas, respectively.
Comparison with 1 mol H2O(g) at 25°C
Let's compare the reaction equilibrium of 1 mole of hydrogen gas and 0.5 mole of oxygen gas with the reaction equilibrium of 1 mole of water vapor at the same temperature.
When we have 1 mole of water vapor (H2O) at 25°C, the reaction will shift backward to reduce the concentration of water vapor and increase the concentration of hydrogen gas and oxygen gas. This is because the reaction is reversible, and the system will try to reach equilibrium by decreasing the concentration of the product (water vapor) and increasing the concentration of the reactants (hydrogen gas and oxygen gas).
Conclusion
In conclusion, the reaction equilibrium of 1 mole of hydrogen gas and 0.5 mole of oxygen gas at 25°C will reach a state where the rates of forward and reverse reactions are equal. The equilibrium constant (Kc) will determine the concentrations of the reactants and the product at equilibrium. By comparing it with the reaction equilibrium of 1 mole of water vapor at the same temperature, we can see that the reaction will shift backward to reduce the concentration of water vapor and increase the concentration of hydrogen gas and oxygen gas.
