At 800 K,the equilibrium constant,K_p,for the following reaction is 3.2 × 10^-7. 2 H₂S(g) 2 H₂(g) + S₂(g)
A reaction vessel at 800 K initially contains 3.00 atm of H₂S.If the reaction is allowed to equilibrate,what is the equilibrium pressure of S₂?

Question

At 800 K,the equilibrium constant,K_p,for the following reaction is 3.2 × 10^-7. 2 H₂S(g)

At 800 K,the equilibrium constant,K<sub>p</sub>,for the following reaction is 3.2 × 10<sup>-7</sup>. 2 H<sub>2</sub>S(g) 2 H<sub>2</sub>(g) + S<sub>2</sub>(g) A reaction vessel at 800 K initially contains 3.00 atm of H<sub>2</sub>S.If the reaction is allowed to equilibrate,what is the equilibrium pressure of S<sub>2</sub>? A) 8.5 × 10<sup>-5</sup> atm B) 6.2 × 10<sup>-3</sup> atm C) 9.0 × 10<sup>-3</sup> atm D) 1.1 × 10<sup>-2</sup> atm E) 1.4 × 10<sup>-2</sup> atm

2 H₂(g) + S₂(g)
A reaction vessel at 800 K initially contains 3.00 atm of H₂S.If the reaction is allowed to equilibrate,what is the equilibrium pressure of S₂?

Quizplus · Accepted Answer

Using the given equilibrium constant: K_p = [S₂] / ([H₂S]^/2 [H₂]) Plugging in the initial pressure of H₂S (3.00 atm) and assuming x is the change in pressure for both S₂ and H₂ gases at equilibrium: K_p = (x)^/2 / (3.00 - x)(3.00 - x) Solving for x using the quadratic formula: x = 8.98 × 10^-4 atm Adding x to the initial pressure of H₂S to get the equilibrium pressure of S₂: Equilibrium pressure = 3.00 atm + 8.98 × 10^-4 atm = 9.00 × 10^-4 atm Rounding to two significant figures gives the final answer of 9.0 × 10^-3 atm, which is choice C.

At 800 K,the Equilibrium Constant,Kp,for the Following Reaction Is 3

At 800 K,the Equilibrium Constant,K_p,for the Following Reaction Is 3