Hydrogen sulfide burns in oxygen to form sulfur dioxide and water. If sulfur dioxide is being formed at a rate of 0.19 mol L-1 s-1, what are the rates of disappearance of hydrogen sulfide and oxygen?
rate of disappearance of O2?
rate of disappearance of H2S?
I got :
rate of disappearance of H2S = 0.19 mol L-1 S-1
rate of disappearance of O2 = 0.29 mol L-1 S-1
but when I checked the answer, it turned out wrong answer.
Am I doing something wrong?
For O2 I would have rounded 0.285 to 0.28 and I think both should have a - sign. Personally, I think the words "rate of disappearance" should take care of the negative sign but sometimes that isn't so.
To determine the rates of disappearance of hydrogen sulfide (H2S) and oxygen (O2), we need to use the balanced equation for the combustion of hydrogen sulfide:
2H2S + 3O2 --> 2SO2 + 2H2O
From the balanced equation, we can see that for every 2 moles of hydrogen sulfide reacted, 2 moles of sulfur dioxide and 2 moles of water are formed. Similarly, for every 3 moles of oxygen reacted, 2 moles of sulfur dioxide and 2 moles of water are formed.
Using stoichiometry, we can relate the rates of disappearance of H2S and O2 to the rate of formation of SO2. The coefficient of H2S in the balanced equation is 2, and the coefficient of O2 is 3.
Rate of disappearance of H2S / 2 = Rate of disappearance of O2 / 3 = Rate of formation of SO2
Given that the rate of formation of SO2 is 0.19 mol L-1 s-1, we can use this information to calculate the rates of disappearance of H2S and O2.
Rate of disappearance of H2S = (0.19 mol L-1 s-1) * 2 / 2 = 0.19 mol L-1 s-1
Rate of disappearance of O2 = (0.19 mol L-1 s-1) * 3 / 2 = 0.285 mol L-1 s-1
So, the correct rates of disappearance are:
Rate of disappearance of H2S = 0.19 mol L-1 s-1
Rate of disappearance of O2 = 0.285 mol L-1 s-1
Therefore, your calculation of the rate of disappearance of H2S is correct, but the calculation for the rate of disappearance of O2 is incorrect.