Hi, I have a question regarding the following chemical equilibrium equation:
2CrO4^2- + 2H3O^+ <--> Cr2O7^2- + 3H2O
The question is: how would you manipulate the above equation to produce more Cr2O7^2- ions without adding any Chromium based compounds? I understand the general principle: increasing the acidity of the solution will drive the reaction to the right, thus producing more dichromate (Cr2O7^2-). Therefore I would have to increase the concentration of H3O. But how would the final balanced equation with increased H3O look like? If someone could show me how to get that equation, i would greatly appreciate that.
Thanks,
Constantine
I don't know what it is you don't understand. Frankly you seem to understand it very well. You add H3O^+ and the rxn shifts to the right producing more dichromate which is what you want. The equation is already there. That is the initial equation, the intermediate equation, the final equation. etc. That IS the equation. You don't need anything more. But something tells me there is a nagging in your mind; tell us what it is so we can help straighten it out to your satisfaction.
Thank you Dr.Bob. I got it now.
To produce more Cr2O7^2- ions without adding any chromium-based compounds, you can indeed increase the acidity of the solution as you mentioned. This can be done by adding a strong acid, such as sulfuric acid (H2SO4), to the equilibrium.
To balance the equation with increased H3O+, you need to add H3O+ ions to the reactant side until the number of hydrogen atoms is equal on both sides of the equation. Here's how you can do it step-by-step:
1. Start with the given equation:
2CrO4^2- + 2H3O^+ <--> Cr2O7^2- + 3H2O
2. To increase the concentration of H3O+, you can add H3O^+ to the reactant side. Since there are 2 H3O^+ ions on the reactant side, you can add 2 H3O^+ ions to balance the equation.
3. After adding 2 H3O^+ ions, the equation becomes:
2CrO4^2- + 2H3O^+ + 2H3O^+ <--> Cr2O7^2- + 3H2O
4. Simplifying the equation, it becomes:
2CrO4^2- + 4H3O^+ <--> Cr2O7^2- + 3H2O
Now, the equation is balanced and shows the increased concentration of H3O^+ ions on the reactant side. By increasing the acidity of the solution, this equilibrium can be driven to the right, favoring the formation of more Cr2O7^2- ions.
To manipulate the given equilibrium equation to produce more Cr2O7^2- ions without adding any Chromium based compounds, you will need to increase the concentration of H3O^+ ions in the system. Here's how you can do it:
Step 1: Start with the original balanced equation:
2CrO4^2- + 2H3O^+ <--> Cr2O7^2- + 3H2O
Step 2: To increase the concentration of H3O^+ ions, you can add a strong acid to the system that will ionize and contribute H3O^+ ions. In this case, we can add HCl (hydrochloric acid) to the reaction.
Step 3: The balanced equation for the addition of HCl will look like this:
HCl + H2O <--> H3O^+ + Cl^-
Step 4: Now, you need to combine the original equation with the equation for the addition of HCl. However, before combining the equations, it is important to note that the overall charge on both sides of the equation should be balanced.
To achieve this, multiply the original equation by 2 and multiply the HCl equation by 1, so the H3O^+ and the OH^- (product of the water dissociation) ions cancel out:
2CrO4^2- + 4H3O^+ <--> 2Cr2O7^2- + 6H2O
HCl + H2O <--> H3O^+ + Cl^-
Step 5: Combine the two equations by adding them together, ensuring that the reactants of both equations appear on the left side and the products appear on the right side:
2CrO4^2- + 4H3O^+ + HCl <--> 2Cr2O7^2- + 6H2O + Cl^-
The final balanced equation, after adding HCl to increase the concentration of H3O^+ ions, will be:
2CrO4^2- + 4H3O^+ + HCl <--> 2Cr2O7^2- + 6H2O + Cl^-
Now, you have successfully manipulated the equation to increase the concentration of Cr2O7^2- ions by adding HCl without introducing any additional chromium-based compounds.