Question

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In A3(g) = 3A (g) reaction, the initial
concentration of A3 is "a" mol L-1 If x is degree of
dissociation of A3. The total number of moles at
equilibrium will be​

Answer 1

Answer:

Given reaction is A3(g)=3A(g)

Initially the number of moles of A is "a"

X is the degree of dissociation

It means xa moles have dissociated so

$A_{3} <=>3A$

at equilibrium condition (a)(1-x)<=>3ax

Total number of moles are a-ax+3ax

                                          =a+2ax

Answer 2

Explanation:

Given reaction is A3(g)=3A(g)

Initially the number of moles of A is "a"

X is the degree of dissociation

It means xa moles have dissociated so

A_{3} < = > 3AA

3

<=>3A

at equilibrium condition (a)(1-x)<=>3ax

Total number of moles are a-ax+3ax

=a+2ax

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