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Standard VII

Chemistry

Zero Order Reaction

The rate cons...

Question

The rate constant of the reaction $A \to B$ is $k = 0.5$ ; the initial concentration of $A$ being $1$ $m o l / l i t r e$ . Calculate the degree of conversion of substance ' $A$ ' within $1 h o u r$ if the reaction is of zeroth, first and second order. How does the degree of conversion depend on the order of reaction?

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Solution

Zero order:
$x = k t$
$x = 0.5 \times 1 = 0.5$
First order: $k = \frac{2.303}{t} {log}_{10} \frac{a}{a - x}$
$k = 0.5;$ $t = 1$ $h r;$ $a = 1$ $M$
$0.5 = \frac{2.303}{1} {log}_{10} \frac{1}{(1 - x)}$
$x = 0.39$
Second order: $k = \frac{1}{t} \frac{x}{a (a - x)}$
$0.5 = \frac{1}{1} \cdot \frac{x}{1 (1 - x)}$
$x = 0.33$

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The rate constant of the reaction A→B is k=0.5; the initial concentration of A being 1 mol/litre. Calculate the degree of conversion of substance 'A' within 1 hour if the reaction is of zeroth, first and second order. How does the degree of conversion depend on the order of reaction?

Zero order:x=ktx=0.5×1=0.5First order: k=2.303tlog10aa−xk=0.5; t=1 hr; a=1 M0.5=2.3031log101(1−x)x=0.39Second order: k=1txa(a−x) 0.5=11⋅x1(1−x) x=0.33