Write the equilibrium constant for this reaction in terms of the equilibrium constants, ka and kb, for reactions a and b below: a.) 2 s(s) 3 o2(g) 2 so3(g) ka b.) s(s) o2(g) so2(g) kb

(a) Ka = \frac{(SO3)^{2} }{(S)^{2}*(O2)^{3}  }

(b) Kb = \frac{(SO2)^{1} }{(S)^{1} *(O2)^{1} }

Explanation:

Correct equation: 2S(s) + 3O_{2}(g) ⇒ 2SO_{3}(g)\\

S(s) + O_{2}(g) ⇒ 2SO_{2}(g)\\

Equilibrium constant refers to a relation between products and reactants of an equilibrium reaction with respect to the components concentration and coefficients.

generally in an equilibrium reaction between reactant A, B, C and D shown below

aA+bB ⇒ cC + dD

the equilibrium constant is represented by

k = \frac{C^{c} *D^{d} }{A^{a} *B^{b} }

comparing this with the equation in the problem the equilibrium constants ka and kb are;

Ka = \frac{(SO3)^{2} }{(S)^{2}*(O2)^{3}  }

Kb = \frac{(SO2)^{1} }{(S)^{1} *(O2)^{1} }

im pretty sure its b

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol