Consider the following reaction:
2H2S+SO2=2H20+S(s)
A reaction mixture initially containing 0.500M H2S and 0.500M SO2 was found to contain 1.0×10−3M at a certain temperature. A second reaction mixture at the same temperature initially contains H2S 0.255M and SO2 0.320M .
Consider the following reaction:
2H2S+SO2=2H20+S(s)
A reaction mixture initially containing 0.500M H2S and 0.500M SO2 was found to contain 1.0×10−3M WHAT at a certain temperature. A second reaction mixture at the same temperature initially contains H2S 0.255M and SO2 0.320M .needs to be finished. If you are trying to cut and paste, please be advised that you must type it. Can't cut and post most things.
To determine the concentrations of the reactants and products in the second reaction mixture, we can use the concept of stoichiometry and apply the principles of chemical equilibrium.
First, let's define the initial concentrations of H2S and SO2 in the second reaction mixture:
[H2S]initial = 0.255 M
[SO2]initial = 0.320 M
We know that the reaction follows the stoichiometry:
2H2S + SO2 -> 2H2O + S
From the balanced equation, we can see that for every 1 mole of SO2, we need 2 moles of H2S to react. Thus, the mole ratio of SO2 to H2S is 1:2.
Now, let's determine the limiting reactant in the second reaction mixture. To do this, we compare the mole ratio of SO2 to H2S with the actual mole ratio of the reactants:
Mole ratio of SO2 to H2S = 1:2
Actual mole ratio (in the second reaction mixture) = [SO2]initial / [H2S]initial
= 0.320 M / 0.255 M
≈ 1.255:1
Since the actual mole ratio of SO2 to H2S is greater than the stoichiometric ratio, it means that H2S is the limiting reactant in this reaction.
To determine the concentration of the remaining SO2 and the products, we need to calculate the amount of H2S reacted. Since we know that H2S is the limiting reactant, it will completely react.
The initial moles of H2S in the second reaction mixture can be calculated as follows:
moles of H2S = [H2S]initial * volume of the mixture
Since we don't have the volume of the mixture, we can assume that the volume is the same for both reaction mixtures. Therefore, the moles of H2S in both reaction mixtures are the same.
Now, let's calculate the concentration of SO2 and the products in the second reaction mixture.
1. Concentration of SO2:
Since H2S is the limiting reactant and all of it reacts, we can assume that the moles of SO2 consumed are equal to half the moles of H2S initially present:
moles of SO2 consumed = moles of H2S (initial)
The moles of SO2 remaining in the second reaction mixture can be calculated as follows:
moles of SO2 remaining = moles of SO2 (initial) - moles of SO2 consumed
Finally, the concentration of SO2 in the second reaction mixture can be calculated as:
[SO2] = moles of SO2 remaining / volume of the mixture
2. Concentration of H2O and S:
Since the reaction is 1:1 for SO2 to H2O, the moles of H2O formed will be equal to the moles of SO2 consumed.
The concentration of H2O and S can be calculated as:
[H2O] = moles of H2O / volume of the mixture
[S] = moles of S / volume of the mixture
Note that the volume of the mixture is assumed to be the same in both reaction mixtures.
To summarize, to determine the concentrations of SO2, H2O, and S in the second reaction mixture, use the following steps:
1. Calculate the moles of H2S initially present in both reaction mixtures.
2. Determine the moles of SO2 consumed by assuming it is half the moles of H2S initially present.
3. Calculate the moles of SO2 remaining in the second reaction mixture.
4. Calculate the concentrations of SO2, H2O, and S by dividing the respective moles by the volume of the mixture.
These calculations will give you the concentrations of the reactants and products in the second reaction mixture.