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

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) = ?

To understand why these problems were solved the way they were, you need to look at the balanced chemical equations and use the stoichiometry of the reactions.

1. Rate of appearance and disappearance:

In the first problem, the balanced equation is:
2 K2Cr2O7 + 8 H2SO4 + 3 C2H6O → 2 Cr(SO4)3 + 2 K2SO4 + 11 H2O

From the balanced equation, you can see that 3 moles of C2H6O react to form 2 moles of Cr(SO4)3. The rate of appearance of Cr(SO4)3 is given as 1.24 mol/min. To calculate the rate of disappearance of C2H6O, you need to use the stoichiometry:

1.24 mol/min Cr(SO4)3 = 2 mol C2H6O / 3 mol Cr(SO4)3 * x mol/min C2H6O

Solving this equation, you get:
x mol/min C2H6O = (1.24 mol/min Cr(SO4)3) * (3 mol Cr(SO4)3 / 2 mol C2H6O) ≈ 1.86 mol/min

So, the rate of disappearance of C2H6O is calculated to be approximately 1.86 mol/min.

2. Rate of production:

In the second problem, the balanced equation is:
2 N2O5 → 4 NO2 + O2

From the balanced equation, you can see that 2 moles of N2O5 react to form 4 moles of NO2. The rate of disappearance of N2O5 is given as 3.7 × 10^-5 mol/min. To calculate the rate of production of NO2, you need to use the stoichiometry:

3.7 × 10^-5 mol/min N2O5 = 4 mol NO2 / 2 mol N2O5 * y mol/min NO2

Solving this equation, you get:
y mol/min NO2 = (3.7 × 10^-5 mol/min N2O5) * (4 mol NO2 / 2 mol N2O5) ≈ 7.4 × 10^-5 mol/min

So, the rate of production of NO2 is calculated to be approximately 7.4 × 10^-5 mol/min.

To understand the rate of disappearance and appearance in chemical reactions, you need to rely on stoichiometry, which involves balancing chemical equations and using the coefficients to relate the quantities of reactants and products.

In the first problem, the balanced equation is:

2 K2Cr2O7 + 8 H2SO4 + 3 C2H6O -> 2 Cr(SO4)3 + 2 K2SO4 + 11 H2O

From the equation, you can see that 3 moles of C2H6O are required to produce 2 moles of Cr(SO4)3. The rate of appearance of Cr(SO4)3 is given as 1.24 mol/min. To find the rate of disappearance of C2H6O, you can set up the following ratio:

(3 moles of C2H6O / 2 moles of Cr(SO4)3) = (rate of disappearance of C2H6O / 1.24 mol/min)

Simplifying this equation, you get:

(rate of disappearance of C2H6O) = (3/2) × (1.24 mol/min) = 1.86 mol/min

Therefore, the rate of disappearance of C2H6O is 1.86 mol/min.

In the second problem, the balanced equation is:

2 N2O5 -> 4 NO2 + O2

From the equation, you can see that 2 moles of N2O5 produce 4 moles of NO2. The rate of disappearance of N2O5 is given as 3.7 × 10^-5. To find the rate of production of NO2, you can use the following ratio:

(2 moles of N2O5 / 4 moles of NO2) = (rate of disappearance of N2O5 / rate of production of NO2)

Simplifying this equation, you get:

(rate of production of NO2) = 2 × (3.7 × 10^-5) = 7.4 × 10^-5

Therefore, the rate of production of NO2 is 7.4 × 10^-5.