Question

for the reaction A+3B=2C+D , initial moles of a is twice that of b. if at equilibrium moles of b and c are equal then % of b reacted is

Answer 1

The given reaction is;

                 $A + 3B$        ⇄      $2C +D$

$t = 0,$         $(a)$      $(a/2)$               $0$       $0$          

$t = eqm,$   $(a-x)$  $(a/2-3x)$     $2x$     $x$

Now, at equilibrium;

$\frac{a}{2} - 3x = 2x$  [Given]

⇒ $\frac{a}{2} = 5x$

⇒ $x = \frac{a}{10}$

∴ Total amount of B reacted = $3x = \frac{3a}{10}$

Ans, initial amount of B = $\frac{a}{2}$

∴ Percentage of B reacted = $[\frac{3a}{10}$ ÷ $\frac{a}{2}]$ × $100$% = $\frac{6}{10}$ × $100$% = 60%

Ans) The percentage of B reacted in 60%

Answer 2

Answer:

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