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Rate Constant

When the rate...

Question

When the rate is determined by the change in concentration of two different reactants, then the kinetic
the equation may be expressed as:

k_{2} = \frac{2.303}{(a - b) t} l o g \frac{(a - x) b}{(b - x) a}

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k_{2} = \frac{2.303}{(a - b) t} l o g \frac{(a - x)}{(a - x)}

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k_{2} = \frac{2.303}{(a - b)} l o g \frac{(a - x)}{(b - x)}

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(D)

k_{2} = \frac{1}{t} \times \frac{x}{(a - x)}

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Solution

The correct option is A $k_{2} = \frac{2.303}{(a - b) t} l o g \frac{(a - x) b}{(b - x) a}$
In a reactions rate is determined by the change in concentration of two different reactants, then it is second order reaction.

For second order reactions kinetic equation expressed as

$k_{2} = \frac{2.303}{(a - b) t} l o g \frac{(a - x) b}{(b - x) a}$

$k_{2}$ is rate constant.

So answer is A.

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When the rate is determined by the change in concentration of two different reactants, then the kineticthe equation may be expressed as:

When the rate is determined by the change in concentration of two different reactants, then the kinetic
the equation may be expressed as: