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Question

For the reaction $$A + 3B \rightleftharpoons 2C + D$$, initial mole of $$A$$ is twice that of $$B$$. If at equilibrium moles of $$B$$ are equal, then percent of $$B$$ reacted is:



A
10%
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B
20%
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C
40%
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D
60%
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Solution

The correct option is D 60%
Given reaction is
                     $$A\quad +\quad 3B\quad  \rightleftharpoons \quad 2C\quad +\quad D$$
$$t=0$$,            $$a$$             $$a/2$$                 $$0$$                $$0$$
$$t=eqm$$,    $$a-x$$   $$a/2-3x$$            $$2x$$              $$x$$
At equilibrium,
$$\cfrac {a}{2}-3x=2x$$ (Given)
$$\cfrac {a}{2}= 5x \Rightarrow x=\cfrac {a}{10}$$
$$\therefore$$ Amount of $$B$$ reacted= $$3x=\cfrac {3a}{10}$$
Initial amount of $$B=a/2$$
So, % of $$B$$ reacted=$$ \left(\cfrac {3a}{10}/\cfrac {a}{2}\right)\times 100$$
                                $$=\cfrac {6}{10}\times 100= 60$$%

Chemistry

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