Answer to Question #108008 in General Chemistry for Beverlie

Question

1.The value of KC for the thermal decomposition of hydrogen sulfide, shown below, is 2.2 × 10-4 at 1400 K.
2H2S(g) ⇌ 2H2(g)+ S2(g)
A sample of gas in which [H2S] = 3.00 M is heated to 1400 K in a sealed vessel. After chemical equilibrium has been achieved, what is the value of [H2S]? Assume no H2 or S2 was present in the original sample. answer in M

Answer 1

Kc = ${\frac {[H_2]^2[S_2]} {[H_2S]^2}}$

Initial concentration of H₂S is 3.00 M. If 2x mol of H₂S were involved in reaction, 2x mol of H₂ and x mol of S₂ were produced. Then in equilibrium state we have:

x mol S₂

2x mol H₂

3.00 - 2x mol H₂S

Kc = ${\frac {[H_2]^2[S_2]} {[H_2S]^2}}={\frac {(2x)^2(x)} {(3-2x)^2}}=2.2*10^{-4}$

$4x^3 = 2.2*10^{-4}*(3-2x)^2=$$2.2*10^{-4}*(9+4x^2-12x)=$$=1.98*10^{-3}+8.8*10^{-4}x^2-2.64*10^{-3}x$

$4x^3-8.8*10^{-4}x^2+2.64*10^{-3}x-1.98*10^{-3}=0$

After solving the equation (i did it on computer) we got x = 0.0764, then amount of H₂S involved in reaction is 2x = 2*0.0764 = 0.1528, in equilibrium [H₂S] = 3 - 0.1528 = 2.8472 M