Consider the following reactions: 1. 2 SO3(g) ⇄ 2 SO2(g) + O2(g) K 1 = 2.3 × 10-7 2. 2 NO3(g) ⇄ 2 NO2(g) + O2 (g) K 2 = 1.4× 10-3 Calculate K for the reaction: SO2(g) + NO3(g) ⇄ SO3(g) + NO2(g)

K =78

Explanation:

Step 1: Data given

2SO3(g) <--> 2SO2(g) + O2(g)  kc = 2.3 x 10^-7

2NO3(g) <--> 2NO2(g) + )2(g)   kc = 1.4 x 10^-3

Step 2: Calculate K

Lets write out the two reactions in the proper order and look at how they sum together:

2 SO2(g) + O2(g)  <---> 2 SO3(g)   (1)

2NO3(g) <---> 2 NO2(g)  + O2(g)    (2)

The two reactions as now written give us the correct reactants and products on the correct sides of the reaction arrow.

Since we have O2 as both a reactant and a product, we can cancel O2 and are not part of the final overall reaction equation.

Koverall =  Kc1  * Kc2

Because we reversed reaction number 1 this affects its Kc via the following:

Krev  =  1/Kfwd.  

We then replace Kc1 with its value for the reverse direction.

So  Koverall now =   (1/Kfwd) * Kc2

The sum of the two reactions above gives us:

2 SO2(g) +  2 NO3(g)  <--->  2 SO3(g)  + 2 NO(g)  

The problem states to give the K value for the reaction where all the numbers in front of the molecules are (1), and we have (2)'s.  So basically  if we multiply the whole reaction by 1/2 we'll get the final overall equation we want.

1/2  ( 2 SO2(g)  + 2 NO3(g)  <> 2 SO3(g)  + 2 NO(g) )

So Kfinal =  (Koverall)^1/2    

K =  ( 1/Kfwd  *  Kc2)^1/2

K =  ( [1 / 2.3 * 10^-7]   *  1.4 * 10^-3)^1/2

K =78

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.