At 90C the following equilibrium is established H2(g) + S(s) = H2S(g) Kc= 6.8x10^-2
If 0.17 moles of H2 and 0.4 mole of Sulfur are heated in a 1L vessel up to 90C, what will be the partial pressure of H2S at equilibrium?
Dr. Bob, first is 0.32 atm correct? Second, if so, is 0.3 atm a better answer considering significant digits?
I agree with 0.322 atm and I would round that to 0.32 atm. You have two places in Kc and 2 places in 0.17. There is only 1 place in the 0.4 mol S; however, since that is a solid AND it is in excess, it never enters into the calculation so I wouldn't use that to determine the number of s.f. Congrats. You did a good job. I don't know how you solved the problem but did you notice that Kc = Kp and you can do the calculation two ways.
Dr. Bob, It's obvious that you're the boss. Thank-you
Excitement aside, how do you solve the problem using the Kp method?
I do understand Kc=Kp though
To find the partial pressure of H2S at equilibrium, we need to calculate the number of moles of H2S formed using the given amounts of H2 and Sulfur.
1. First, convert the given amounts of H2 and Sulfur to moles:
moles of H2 = 0.17 mol
moles of Sulfur = 0.4 mol
2. Next, use the stoichiometric ratio of the balanced equation to determine the number of moles of H2S formed. From the equation H2(g) + S(s) = H2S(g), the ratio is 1:1. Therefore, the moles of H2S formed will be equal to the lesser of the moles of H2 and Sulfur, which is 0.17 mol.
3. Since the reaction is carried out in a 1L vessel, the number of moles of H2S at equilibrium will be its molar concentration, given by:
[H2S] = moles of H2S / volume of vessel
= 0.17 mol / 1 L
= 0.17 M
4. Finally, we use the equilibrium constant expression, Kc, to calculate the partial pressure of H2S. Since H2S is in the gas phase, its partial pressure can be equated to its molar concentration:
P(H2S) = [H2S]
= 0.17 atm
Regarding the given answers, we have calculated the partial pressure of H2S to be 0.17 atm. Therefore, based on significant digits, 0.2 atm would be a better answer, as it follows the rule of rounding the final digit to the nearest value when the digit to the right is 5 or more.