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For the reaction NH(3)(g)hArr(1)/(2)N(...

For the reaction
`NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`
Show that the degree of dissociation of `NH_(3)` is given as
`alpha=[1+(3sqrt(3))/4p/K_(p)]^(-1//2)`
where p is equilibrium pressure. If `K_(p)` of the above reaction is `78.1 atm` at `400^(@)C`, calculate `K_(c )`.

Text Solution

AI Generated Solution

To solve the problem, we will follow these steps systematically: ### Step 1: Write the balanced chemical equation The reaction given is: \[ \text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \] ### Step 2: Define the degree of dissociation Let \( \alpha \) be the degree of dissociation of ammonia (\( \text{NH}_3 \)). Initially, we have 1 mole of \( \text{NH}_3 \) and no products. At equilibrium, the dissociation can be represented as: ...
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