All (g)
3N2H4 + 4ClF3 <=> 12HF + 3N2 + 2Cl2
The mixture initially consisted of 0.880 atm N2H4 and 0.970 atm CIF3(g). At equilibrium the partial pressure of N2 is 0.525 atm. Calculate partial pressure of HF at equilibrium
To calculate the partial pressure of HF at equilibrium, we need to use the given initial pressure values and the balanced chemical equation.
Let's first write down the balanced chemical equation:
3N2H4 + 4ClF3 <=> 12HF + 3N2 + 2Cl2
Now, let's analyze the stoichiometry of the equation to understand the relationship between the partial pressures of the different gases.
From the balanced equation, we can see that for every 3 moles of N2H4, we get 12 moles of HF. This means that the ratio between their partial pressures will be the same as the stoichiometric coefficients: 12:3, or 4:1.
Now, let's use this ratio to find the partial pressure of HF. First, we need to determine the partial pressure of N2H4 at equilibrium.
The given initial partial pressure of N2H4 is 0.880 atm. At equilibrium, the partial pressure of N2 is 0.525 atm. Since N2H4 and N2 have a stoichiometric ratio of 1:1, the partial pressure of N2H4 at equilibrium will also be 0.525 atm.
Now, we can use this partial pressure of N2H4 to find the partial pressure of HF. According to the stoichiometric ratio of 4:1, the partial pressure of HF will be:
(4/1) * (0.525 atm) = 2.10 atm
So, the partial pressure of HF at equilibrium is 2.10 atm.