When I invert the concentration of say this equation:
2HCr4- + 3HS03- + 5H+ -> 2Cr3+ + 3SO4^2- + 5H20.
and get RSRRE wrt HS03-, the rates of reactants is positive and the rates of products is negative.
If say it is wrt HS03- which gives it a coefficient of 1 and you are trying to get the ratee of say Cr3+, would the rate be changed back to a positive number? Or will it stay negative?
RSRRE = Relative stoichiometric rxn rate expression.
TIA
To understand the change in sign of the rate expression when you invert the concentration of a reactant, let's break down the process step by step.
In the given reaction:
2HCr4- + 3HSO3- + 5H+ -> 2Cr3+ + 3SO4^2- + 5H2O
The rate expression, taking into consideration the stoichiometry, can be written as follows:
Rate = k [HCr4-]^2 [HSO3-]^3 [H+]^5
Now, if you invert the concentration of HSO3-, the new rate expression would be:
Rate' = k [HCr4-]^2 [1/HSO3-]^3 [H+]^5
Simplifying this expression gives:
Rate' = k [HCr4-]^2 / [HSO3-]^3 [H+]^5
Notice that even after inverting [HSO3-], the rate expression for Cr3+ still has a negative sign in the numerator. Therefore, the rate of Cr3+ would remain negative. The reason for this is that the stoichiometric coefficient of Cr3+ (2) is positive, while the stoichiometric coefficient of HSO3- (3) is also positive, resulting in a negative sign in the rate expression.
Hence, in this specific case, the rate of Cr3+ would still be negative even after inverting the concentration of HSO3-.