Enter your answer in the provided box. Hydrogen sulfide decomposes according to

Question

Enter your answer in the provided box.

Hydrogen sulfide decomposes according to the following reaction, for which

Kc = 9.30 × 108 at 700°C:

2 H2S(g) 2 H2(g) + S2(g)

If 0.55 mol of H2S is placed in a 3.0L container, what is the equilibrium concentration of H2(g) at 700°C?

? M Enter your answer in the provided box.

Hydrogen sulfide decomposes according to the following reaction, for which

Kc = 9.30 × 108 at 700°C:

2 H2S(g) 2 H2(g) + S2(g)

If 0.55 mol of H2S is placed in a 3.0L container, what is the equilibrium concentration of H2(g) at 700°C?

? M

Explanation / Answer

The equilibrium concentration of H is 0.012 mol/L.

Calculate the initial concentrations

[HS] = 0.29mol3.0L = 0.096 67 mol/L (2 significant figures + 2 guard digits)

Write the balanced equation and set up an ICE table.

2HS 2H + S
I/mol·L¹: 0.096 67; 0; 0
C/mol·L¹: -2x; +2x; +x
E/mol·L¹: 0.096 67 - 2x; 2x; x

Write the Kc expression and solve for x

Kc=[H]2[S][HS]2 = 9.30 × 10

(2x)2×x(0.09667–2x)2 = 9.30 × 10

Because Kc is so small, we can assume that 2x is negligible compared to 0.09667. Then we have

4x30.096672 = 9.30 × 10

4x3 = 0.096 67² × 9.30 × 10 = 8.691 × 10¹

x3 = 2.173 × 10¹

x = 6.012 × 10

1

[H] = 2x mol/L= 2 × 6.012 × 10 mol/L = 0.001 20 mol/L

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