Question

at 1100 k, ( k_{p}=0.15 ) for the reaction
( 2 mathrm{so}_{2}(g)+mathrm{o}_{2}(g)
ightleftharpoons 2 mathrm{so}_{3}(g) )
what is the value of ( k ) at this temperature?
( k= )

Explanation:

Step1: Recall the relationship between \(K_p\) and \(K\)

The relationship is \(K_p = K(RT)^{\Delta n}\), where \(\Delta n\) is the change in the number of moles of gas.
For the reaction \(2SO_2(g)+O_2(g)
ightleftharpoons 2SO_3(g)\), \(\Delta n=n_{products}-n_{reactants}=(2)-(2 + 1)=- 1\).
The gas constant \(R = 0.0821\space L\cdot atm/(mol\cdot K)\) and \(T = 1100\space K\).

Step2: Rearrange the formula to solve for \(K\)

From \(K_p = K(RT)^{\Delta n}\), we can solve for \(K\) as \(K=\frac{K_p}{(RT)^{\Delta n}}\).
Substitute the values: \(K=\frac{0.15}{(0.0821\times1100)^{-1}}\).
First, calculate \(0.0821\times1100 = 90.31\).
Then \((0.0821\times1100)^{-1}=\frac{1}{90.31}\).
So \(K = 0.15\times90.31\).

Step3: Calculate the value of \(K\)

\(K=0.15\times90.31 = 13.5465\approx14\)

Answer:

\(14\)