Which is a correct integrated rate equation?

k = - \frac{2.303}{t} l o g \frac{a}{a - x}

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k = \frac{- 2.303}{t} l o g \frac{a - x}{a}

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- d (a - x) = k d t

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All are integrated rate equations

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Solution

The correct option is C $k = \frac{- 2.303}{t} l o g \frac{a - x}{a}$
The integrated rate equation for zero order reaction is $t = \frac{x}{k}$
and for first order reaction is $k = \frac{2.303}{t} l o g \frac{a}{a - x}$
or $= - \frac{2.303}{t} l o g \frac{(a - x)}{a}$

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Similar questions

Q. If a reaction obeys the following equation

k = \frac{2.303}{t} l o g \frac{a}{a - x}

then the order is:

If [A] is the concentration of reactant A at any time t and [A_o] is the concentration at t=0, then for the first order reaction the rate equation can be written as:

Q. Rate constant k varies with temperture by equation

l o g k (m i n^{1}) = l o g 5 \frac{2000 k c a l}{R T \times 2.303} .

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Which is a correct integrated rate equation?

The correct option is C k=−2.303tloga−xaThe integrated rate equation for zero order reaction is t=xkand for first order reaction is k=2.303tlogaa−xor =−2.303tlog(a−x)a

The correct option is C $k = \frac{- 2.303}{t} l o g \frac{a - x}{a}$
The integrated rate equation for zero order reaction is $t = \frac{x}{k}$
and for first order reaction is $k = \frac{2.303}{t} l o g \frac{a}{a - x}$
or $= - \frac{2.303}{t} l o g \frac{(a - x)}{a}$