Asked by Anonymous
Can anyone explain to me the rate of disappearance and appearance? I had two problems and ive figured them out but i don't understand why they were solved this way.
If the rate of appearance of Cr(SO4)3 is 1.24mol/min then what is the disappearance of C2H6O?
2 K2Cr207 + 8 H2SO4 + 3C2H6O -> 2 Cr(SO4)3 + 2 K2S04 + 11 H20
I solved it by 3/2 = 1.5 * 1.24 mol/min = 1.86 mol/min
What is the rate production of NO2 given the disappearance of N2O5 is 3.7 * 10^-5
2 N2O5 -> 4 NO2 + O2
I solved it by 2(3.7 *10^-5) = 7.4*10^-5
If the rate of appearance of Cr(SO4)3 is 1.24mol/min then what is the disappearance of C2H6O?
2 K2Cr207 + 8 H2SO4 + 3C2H6O -> 2 Cr(SO4)3 + 2 K2S04 + 11 H20
I solved it by 3/2 = 1.5 * 1.24 mol/min = 1.86 mol/min
What is the rate production of NO2 given the disappearance of N2O5 is 3.7 * 10^-5
2 N2O5 -> 4 NO2 + O2
I solved it by 2(3.7 *10^-5) = 7.4*10^-5
Answers
Answered by
DrBob222
It works the same way as stoichiometry.
rate Cr(SO$)3 x (3 mol C2H6O/2 mol Cr(SO4)3) = 1.24 x 3/2 = ?
Note that Cr(SO4)3 cancels but C2H6O stays to change the units to that specie.The coefficients do the job.
Same for the other problem.
3.7E-5 x (4/2) = ?
rate Cr(SO$)3 x (3 mol C2H6O/2 mol Cr(SO4)3) = 1.24 x 3/2 = ?
Note that Cr(SO4)3 cancels but C2H6O stays to change the units to that specie.The coefficients do the job.
Same for the other problem.
3.7E-5 x (4/2) = ?
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