1. Consider the following reaction: SO_2Cl_2(g) <=> SO_2(g)+Cl_2(g) K_p = 2.91x1

Question

1. Consider the following reaction:
SO_2Cl_2(g) <=> SO_2(g)+Cl_2(g)
K_p = 2.91x10^3 at 298K
In a reaction at equilibrium, the partial pressure of SO_2 is 116 torr and that of Cl_2 is 261 torr.
What is the partial pressure of SO_2Cl_2 in this mixture?
P=_____torr


2. Consider the following reaction:
NH_4HS(s) <=> NH_3(g)+H_2S(g)
An equilibrium mixture of this reaction at a certain temperature was found to have [NH_3]=0.278M and [H_2S]=0.360M.
What is the value of the equilibrium constant (K_c) at this temperature?
K_c=_________


3. The reaction below has an equilibrium constant:
K_p=2.2x10^6 at 298 K
2COF_2(g) <=> CO_2(g)+CF_4(g)

Calculate K_p for the reactions below:
a. COF_2(g) <=>(1/2)CO_2(g)+(1/2)CF_4(g)
and
b. (2/3)COF_2(g) <=> (1/3)CO_2(g)+(1/3)CF_4(g)
and
c. 2CO_2(g)+2CF_4(g) <=> 4COF_2(g)

Explanation / Answer

1] SO2Cl2 (g) SO2 (g) + Cl2 (g) Kp = P(SO2) X P(Cl2) / P(SO2Cl2) where P(SO2) is the equilibrium partial pressure of SO2 = 110 torr P(Cl2) is the equilibrium partial pressure of Cl2 = 259 torr P(SO2Cl2) is the equilibrium partial pressure of SO2Cl2 = ? and Kp is given as 2.91 X 10^3 putting the values.... 2.91 X 10^3 = 110 X 259 / P(SO2Cl2) 2910 = 28490 / P(SO2Cl2) P(SO2Cl2) = 28490/2910 = 9.790 torr so partial pressure of SO2Cl2 at equilibrium is 9.790 torr 3 You have to figure out the pattern by looking at the coefficients. Since reaction (a) has half the coefficients of the original reaction, just divide the equilibrium constant by 2. For part (b) the coefficients are 3 times the original reaction, multiply the original K by 3. For the last one, it is a reverse reaction so you do the inverse of the original equilibrium constant given to the K^-1 or just 1/k. And that's it

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.