Question

At a certain temperature, the half life period for the catalytic decomposition of ammonia were found to be a follows. Pressure (mm): 100 200 Half life (min.) : 1.92 1.0 What is the unit of rate constant?​

Answer 1

Answer

(t

1/2

)

2

(t

1/2

)

1

=(

a

1

a

2

)

n−1

where, n is order of reaction

From the given data,

1.92

3.52

=(

6667

13333

)

n−1

(a∝ initial pressure)

=(2)

n−1

log

1.92

3.52

=(n−1)log2=0.3010×(n−1)

0.2632=0.3010×(n−1)

n=1.87≈2

Similar calculations are made between first and third observations, n comes equal to 1.908 (≈2).

Thus, the reaction is of second order.

$Answer</p><p></p><p>Open in answr app</p><p></p><p>(t1/2)2(t1/2)1=(a1a2)n−1   where, n is order of reaction</p><p>From the given data,</p><p>1.923.52=(666713333)n−1         (a∝ initial pressure)</p><p>          =(2)n−1</p><p>log1.923.52=(n−1)log2=0.3010×(n−1)</p><p>0.2632=0.3010×(n−1)</p><p>n=1.87≈2</p><p>Similar calculations are made between first and third observations, n comes equal to 1.908 (≈2).</p><p>Thus, the reaction is of second order.</p><p></p><p></p><p>$

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