The equilibrium constant, Kp, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g) 2CH2Cl2(g) Calculate the equilibrium partial pressures of all species when CH4 and CCl4, each at an intitial partial pressure of 1.06 atm, are introduced into an evacuated vessel at 350 K. PCH4 = atm PCCl4 = atm PCH2Cl2 = atm

pCH4 =  0.9184 atm

pCCl4 = 0.9184 atm

pCH2Cl2 = 0.2832 atm

Explanation:

Step 1: Data given

The equilibrium constant, Kp= 9.52 * 10^-2

Temperature = 350 K

Each have an initial pressure of 1.06 atm

Step 2: The balanced equation

CH4(g) + CCl4(g) ⇆ 2CH2Cl2(g)

Step 3: The pressure at the equilibrium

pCH4 = 1.06 - X atm

pCCl4 = 1.06 - X atm

pCH2Cl2 = 2X

Step 4: Calculate Kp

Kp = (2X)² / (1.06 - X)*(1.06 - X)

9.52 * 10^-2 = 4X² / (1.06 - X)*(1.06 - X)

X = 0.1416

Step 5: Calculate the partial pressure

pCH4 = 1.06 - 0.1416 =  0.9184 atm

pCCl4 = 1.06 - 0.1416 =  0.9184 atm

pCH2Cl2 = 2 * 0.1416 = 0.2832 atm

Kp = (0.2832²) / (0.9184*0.9184)

Kp = 9.52 * 10^-2

pCH4 =  0.9184 atm

pCCl4 = 0.9184 atm

pCH2Cl2 = 0.2832 atm

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.