In the reaction A + B $\to$ C + D, the concentration of A and B are equal and the rate of the reaction is rate = k[A][B]. The intergrated rate equation for this reaction is:

k = \frac{x}{t (a - x)}

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k = \frac{x a}{t (a - x)}

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k = \frac{1}{t} \cdot \frac{x}{a (x - a)}

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k = \frac{1}{t} \cdot \frac{x}{t (a (x - a)}

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Solution

The correct option is C $k = \frac{1}{t} \cdot \frac{x}{a (x - a)}$
SECOND ORDER REACTION
A + A $\to$ Product
A + B $\to$ Product
At $t = 0$ a a $0$
At $t = t$
$(a - x)$ $(a - x)$ $x$
At $t = t_{1}$ $(a - x_{1})$ $(a - x_{1})$ $x_{1}$
At $t = t_{2}$ $(a - x_{2})$ $(a - x_{2})$ $x_{2}$
As per rate law, $\frac{d x}{d t} = k_{2} [A]^{n} = k_{2} [A]^{2} = k_{2} [A] [B]$
$∴ (\frac{d x}{d t}) = k_{2} (a - x)^{2}$ ( $k_{2}$ = rate constant for second order reaction)
Also, $k_{2} = \frac{1}{t} [\frac{1}{(a - x)} - \frac{1}{a}] = \frac{1}{t} \frac{a}{a (a - x)}$ or $k_{2} = \frac{1}{(t_{2} - t_{1})} [\frac{1}{(a - x_{2})} - \frac{1}{(a - x_{1})}]$

Where $(a - x_{1})$ and $(a - x_{2})$ are the concentration of the reactant A at time $t_{1}$ and $t_{2}$ respectively. If reactant $A$ and $B$ have different concentrations $a$ and $b$ , then $k_{2} = \frac{2.3030}{t (a - b)} {log}_{10} \frac{b (a - x)}{a (b - x)}$

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	A	+	A	$\to$	Product
	A	+	B	$\to$	Product
At $t = 0$	a		a		$0$
At $t = t$	$(a - x)$		$(a - x)$		$x$
At $t = t_{1}$	$(a - x_{1})$		$(a - x_{1})$		$x_{1}$
At $t = t_{2}$	$(a - x_{2})$		$(a - x_{2})$		$x_{2}$

In the reaction A + B → C + D, the concentration of A and B are equal and the rate of the reaction is rate = k[A][B]. The intergrated rate equation for this reaction is:

In the reaction A + B $\to$ C + D, the concentration of A and B are equal and the rate of the reaction is rate = k[A][B]. The intergrated rate equation for this reaction is: