Question

iclicker question 2 (3 mins)
to improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the contaminant hydrogen sulfide with o₂: 2h₂s(g) + o₂(g) ⇌ 2s(s) + 2h₂o(g)
which of the following statements, i - iv, is/are true?
i. when o₂ is added, this will cause q > k.
ii. when h₂s is removed, this will cause q > k.
iii. when s is added, this will cause q > k.
iv. if h₂o is removed, this will cause q < k.

Explanation:

Step1: Write the expression for \(Q\) and \(K\)

For the reaction \(2H_{2}S(g)+O_{2}(g)
ightleftharpoons 2S(s)+2H_{2}O(g)\), the reaction - quotient \(Q=\frac{P_{H_{2}O}^{2}}{P_{H_{2}S}^{2}\cdot P_{O_{2}}}\) (since \(S\) is a solid and its activity is \(1\)). At equilibrium \(Q = K\).

Step2: Analyze statement I

When \(O_{2}\) is added, the denominator \(P_{H_{2}S}^{2}\cdot P_{O_{2}}\) increases. So, \(Q=\frac{P_{H_{2}O}^{2}}{P_{H_{2}S}^{2}\cdot P_{O_{2}}}\) decreases. So, \(Q

Step3: Analyze statement II

When \(H_{2}S\) is removed, the denominator \(P_{H_{2}S}^{2}\cdot P_{O_{2}}\) decreases. So, \(Q=\frac{P_{H_{2}O}^{2}}{P_{H_{2}S}^{2}\cdot P_{O_{2}}}\) increases. So, \(Q > K\).

Step4: Analyze statement III

Since \(S\) is a solid, adding \(S\) does not change the reaction - quotient \(Q\) (because the concentration/partial pressure of a solid is considered constant in the expression of \(Q\) and \(K\)). So, \(Q = K\). Statement III is false.

Step5: Analyze statement IV

When \(H_{2}O\) is removed, the numerator \(P_{H_{2}O}^{2}\) decreases. So, \(Q=\frac{P_{H_{2}O}^{2}}{P_{H_{2}S}^{2}\cdot P_{O_{2}}}\) decreases. So, \(Q

Answer:

II and IV are true.