For the reaction
`2HI(g)hArrH_(2)(g)+I_(2)(g)`
The degree of dissociation `(alpha)` of `HI(g)` is related to equilibrium constant `K_(p)` by the expression
a. `(1+2sqrt(K_(p)))/2`, b. `sqrt((1+2K_(p))/2)`
c. `sqrt((2K_(p))/(1+2K_(p)))`, d. `(2sqrt(K_(p)))/(1+2sqrt(K_(p)))`

Explanation

To derive the relationship between the degree of dissociation (α) of HI and the equilibrium constant (Kp) for the reaction:

\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]

we can follow these steps:

### Step 1: Define the Degree of Dissociation
The degree of dissociation (α) is defined as the fraction of the initial moles of a substance that dissociates. For the reaction, if we start with 2 moles of HI, at equilibrium we have:
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