A + B ==> C
K = (C)/(A)(B)
Substitute for C and A and B and solve for K in each instance.
I have 6.515 for trial 1, 6.47 for trial 2, and 6.52 for trial 3 which is very close to the same value for K at each of the three trials. You need to do it yourself with more accuracy and don't forget to round to the correct number of significant figures.
Equilibrium concentrations
(a)(mol/L)// (b)(mol/L)// (c)(mol/L)
0.040 // 0.066 // 1.72x10^-2
0.080 // 0.017 // 8.8x10^-3
0.030 // 0.024 // 4.7x10^-3
My data was collected at 25○C for the reaction A(g) + B(g) = C(g)
What i have to do now is involving three different initial sets of concentrations, use the data to show that K does not vary with initial concentrations at the same time.
I am very stuck on this question i just wish i could understand it more.
3 answers
okay i have the vaule of K being 6.527.
I'm just confused as to how to start the teials. do i just do (b)/(a)(c), (a)/(b)(c) and (c)/(a)(b) ?
I'm just confused as to how to start the teials. do i just do (b)/(a)(c), (a)/(b)(c) and (c)/(a)(b) ?
i got all the answers now Doctor Bob. thank you so much for your help! its muchly appreciated!