Answer
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Hint: In equilibrium condition, the rate of forward reaction is equal to the rate of backward reaction. It is given that the concentration of the reactant A is the same as the concentration of product D at the equilibrium. First we need to determine the change in concentration from initial to final.
Complete step by step answer:
The chemical equilibrium is the state of the system where the concentration of the reactant and the concentration of the product does not change with time.
The system is said to be in equilibrium condition when the rate of forward reaction is equal to the rate of backward reaction.
To determine the concentration of the reactant B, first we need to determine the ICE table which is the initial, change, equilibrium.
The given reaction is shown below.
$A + 2B \rightleftharpoons 2C + D$
Here, one mole of A is reacting with 2 mole of B to give 2 mole of C and one mole of D.
Let's take the initial concentration of A as a and the initial concentration of B is 1.5 times of A.
$A + 2B \rightleftharpoons 2C + D$
We know that at equilibrium the concentration of A and D are the same.
$A = D$
So,
$ \Rightarrow (a - x) = x$
$ \Rightarrow a = 2x$
$ \Rightarrow x = 0.5$
We know the concentration of B at equilibrium is
$ \Rightarrow 1.5(a - 2x)$
Substitute the value of x in the above expression.
$ \Rightarrow 1.5(a - 2 \times 0.5)$
$ \Rightarrow 1.5(a - 1)$
$ \Rightarrow 0.5a$
$ \Rightarrow \dfrac{a}{2}$
Thus, the concentration of B at equilibrium is $\dfrac{a}{2}$.
Therefore, the correct option is B.
Note:
Make sure to draw the ICE table without which we cannot determine the correct expression for calculation the answer. It is the key step for solving the question based on chemical equilibrium.
Complete step by step answer:
The chemical equilibrium is the state of the system where the concentration of the reactant and the concentration of the product does not change with time.
The system is said to be in equilibrium condition when the rate of forward reaction is equal to the rate of backward reaction.
To determine the concentration of the reactant B, first we need to determine the ICE table which is the initial, change, equilibrium.
The given reaction is shown below.
$A + 2B \rightleftharpoons 2C + D$
Here, one mole of A is reacting with 2 mole of B to give 2 mole of C and one mole of D.
Let's take the initial concentration of A as a and the initial concentration of B is 1.5 times of A.
$A + 2B \rightleftharpoons 2C + D$
Initial | a | 1.5a | 0 | 0 |
Final | a-x | 1.5(a-2x) | 2x | x |
We know that at equilibrium the concentration of A and D are the same.
$A = D$
So,
$ \Rightarrow (a - x) = x$
$ \Rightarrow a = 2x$
$ \Rightarrow x = 0.5$
We know the concentration of B at equilibrium is
$ \Rightarrow 1.5(a - 2x)$
Substitute the value of x in the above expression.
$ \Rightarrow 1.5(a - 2 \times 0.5)$
$ \Rightarrow 1.5(a - 1)$
$ \Rightarrow 0.5a$
$ \Rightarrow \dfrac{a}{2}$
Thus, the concentration of B at equilibrium is $\dfrac{a}{2}$.
Therefore, the correct option is B.
Note:
Make sure to draw the ICE table without which we cannot determine the correct expression for calculation the answer. It is the key step for solving the question based on chemical equilibrium.
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