Consider the following balanced thermochemical equation for a reaction sometimes used for
H2S production:
1/8 S8 (s) + H2 (g) �¨ H2S (g) ƒ¢Hrxn = -20.2 kJ
(a)Is this an exothermic or endothermic reaction?
(b)What is ĢHrxn for the reverse reaction?
(c)What is ĢH when 3.9 mol of S8 react?
I know it's exothermic
the reverse reaction is 20.2
but with part c, I am not sure: is this right?
I get 3.9 moles s8 . (-20/2kj/1/8 mol s8 = 630.24 kj
Well, you're almost there with part c! Let's go through it together. First, let's multiply the ΔH value by 3.9 mol S8 to account for the given amount of reactant:
-20.2 kJ/mol * 3.9 mol S8 = -78.78 kJ
But remember, the balanced equation tells us that it takes 1/8 mol S8 to produce the given ΔH value. So, we need to divide by 1/8 to find the ΔH when 3.9 mol S8 reacts:
(-78.78 kJ) / (1/8 mol S8) = -78.78 kJ * (8 mol S8) = -630.24 kJ
So, the ΔH when 3.9 mol of S8 reacts is indeed -630.24 kJ. Good job with the calculations!
Yes, you are correct that the reaction is exothermic based on the given thermochemical equation.
For part (b), to find ΔHrxn for the reverse reaction, you need to reverse the direction of the reaction. So, the balanced reverse reaction would be:
H2S (g) → 1/8 S8 (s) + H2 (g)
The value of ΔHrxn for the reverse reaction would be the opposite sign of the original reaction. Therefore, it would be +20.2 kJ.
For part (c), you would need to use the stoichiometry of the reaction to find the change in enthalpy when 3.9 moles of S8 react. From the balanced equation, you can see that for every 1/8 mole of S8, there is a change in enthalpy of -20.2 kJ.
Using this information, you can set up a proportion:
-20.2 kJ / (1/8 mol S8) = ΔH / (3.9 mol S8)
Simplifying the proportion, you would get:
ΔH = (-20.2 kJ / (1/8 mol S8)) * (3.9 mol S8)
Calculating this, the change in enthalpy (ΔH) when 3.9 moles of S8 react would be -630.24 kJ. Therefore, your calculation is correct.
To determine the value of ĢH when 3.9 mol of S8 react, you first need to calculate the moles of S8 involved in the reaction. According to the balanced equation, the stoichiometry between S8 and H2S is 1:1/8. Therefore, for every one mole of S8, 1/8 moles of S8 are used.
So, to calculate the moles of S8 used when 3.9 mol of S8 react, you can use the following equation:
moles of S8 = (moles of S8 react) x (1/8)
moles of S8 = 3.9 mol x (1/8) = 0.4875 mol S8
Now that you have the moles of S8 involved in the reaction, you can calculate the ĢH using the given ĢH value of -20.2 kJ for the reaction:
ĢH = (moles of S8) x (ĢHrxn / moles of S8 in the balanced equation)
ĢH = 0.4875 mol x (-20.2 kJ / 1/8 mol) = -195.51875 kJ
Therefore, the value of ĢH when 3.9 mol of S8 react is approximately -195.52 kJ.