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

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.