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Hydrogen sulfide decomposes according to the following
Chapter 17, Problem 40P(choose chapter or problem)
Problem 40P
Hydrogen sulfide decomposes according to the following reaction, for which Kc= 9.30× 10?8 at 700°C:
2H2S(g) ⇌ 2H2(g) + S2(g)
If 0.45 mol of H2S is placed in a 3.0-L container, what is the equilibrium concentration of H2(g) at 700°C?
Questions & Answers
QUESTION:
Problem 40P
Hydrogen sulfide decomposes according to the following reaction, for which Kc= 9.30× 10?8 at 700°C:
2H2S(g) ⇌ 2H2(g) + S2(g)
If 0.45 mol of H2S is placed in a 3.0-L container, what is the equilibrium concentration of H2(g) at 700°C?
ANSWER:
Solution 40P:
Here, we are going to calculate the equilibrium concentration of H2(g) at 700°C.
Step 1:
Given reaction,
2H2S(g) ⇌ 2H2(g) + S2(g)
Volume = 3.0 L
Moles of H2S =0.45 mol
Therefore, initial [H2S] = =.0.15 M
Let be the concentration of H2(g) and S2 formed after some time. With change in time the concentration of H2S(g) will be reduced by -2x.
Therefore,
At change, [H2] = , [S2] =
and [H2S] =
Thus,
At equilibrium,
[HI] = , [H2] =
and [S2] =
Hence, the reaction table for the reaction is given below:
