So I have a solution of KNO3 (690g in 1.6L) and I add 50g of K2CrO4 to the solution. I´ll warm the solution up to 75degree celsius and I´ve to determine at which temperature precipitate will...
So I have a solution of KNO3 (690g in 1.6L) and I add 50g of K2CrO4 to the solution. I´ll warm the solution up to 75degree celsius and I´ve to determine at which temperature precipitate will form.
KNO3 (s) <---> K+ (aq) + NO3- (aq)
So I made up an hypothethis that since both KNO3 and K2CrO4 share a common ion then it´ll add up to the concentration of [K+] when I calculate the initial-change-end.
I also think that the reaction will shift to the left, and produce more KNO3 (s), because the solubility will decrease as both reactants have a common ion.
I´ve calculated the temperatur based on equilibrium and thermodynamic knowledge and I got an solution, but the temperature seems weird because I think it´s way too high? From my hypothethis precipitate will form much easier with the K2CrO4 in solution, than without. But the temperature is still way to high?
There are several things that you need to consider in this experiment. First, we need to be sure that the solution is homogenous, meaning that all the KNO3 and K2CrO4 are dissolved in water at 75 degrees Celsius. You don’t actually need to compute first for the temperature, you just have to observe the solution as the temperature is cooled. Second, you actually do not need to add any K2CrO4 in order for the KNO3 to precipitate. By just lowering the temperature, KNO3 crystals will precipitate. This is because at lower temperature, the saturation of the solution will change. This however allows the formation of crystals in the solution.
You can confirm this by plotting the curve to get the temperature. The equation you gave was correct. So if we have
`Delta G = Delta H - T Delta S`
`Delta G = -RT lnQ`
We can have:
`-RT ln Q =Delta H - T Delta S`
where:` ln Q = [K^(+)][NO_3 ^(-)]`
Arranging the equation in the form af a line equation, y = mx + b , we can have:
`ln Q = (-(Delta H)/(R))(1/T) + ((Delta S)/(R)) `
Plot the ln Q vs. 1/T with a slope of `-(Delta H)/(R)` and a y-intercept of `(Delta S)/(R)` .
You should have to test different concentration of potassium nitrate in order to generate different ln Q values and eventually create a linear plot. This experiment is to determine the thermodynamic properties of potassium nitrate so we need the values of temperature in order to get the values of Delta H, G and S at particular temperature.