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3 years ago

A certain first-order reaction 45% complete in 65seconds, determine the rate constant and the half life for the process

Chemistry

1 answer:

masha68 [24]3 years ago

3 0

The rate constant : k = 9.2 x 10⁻³ s⁻¹

The half life : t1/2 = 75.3 s

<h3>Further explanation</h3>

Given

Reaction 45% complete in 65 s

Required

The rate constant and the half life

Solution

For first order ln[A]=−kt+ln[A]o

45% complete, 55% remains

A = 0.55

Ao = 1

Input the value :

ln A = -kt + ln Ao

ln 0.55 = -k.65 + ln 1

-0.598=-k.65

k = 9.2 x 10⁻³ s⁻¹

The half life :

t1/2 = (ln 2) / k

t1/2 = 0.693 : 9.2 x 10⁻³

t1/2 = 75.3 s

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<u>Answer:</u>

<u>For A:</u> The $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u> The $K_c$ for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

<u>For A:</u>

The expression of $K_p$ for the above reaction follows:

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<u>For B:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

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where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

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A certain first-order reaction 45% complete in 65seconds, determine the rate constant and the half life for the process ​

A certain first-order reaction 45% complete in 65seconds, determine the rate constant and the half life for the process