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प्रश्न
Which of the following correctly represents integrated rate law equation for a first order reaction in a gas phase?
विकल्प
k = `2.303/t xx log_10 P_i/(P_i - P)`
k = `2.303/t xx log_10 P_i/(2P_i - P)`
k = `2.303/t xx log_10 (2P_i)/(P_i - P)`
k = `2.303/t xx log_10 (P_i - P)/(2P_i)`
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उत्तर
`bb(k = 2.303/t xx log_10 P_i/(2P_i - P))`
Explanation:
For the gas phase reaction,
\[\ce{A(g) -> B(g) + C(g)}\]
Let initial pressure of A be Pi that decreases by x within time t.
Pressure of reactant A at time t
PA = Pi – x .....(i)
The pressures of products B and C at time t are
PB = PC = x
The total pressure at time t is then
P = Pi – x + x + x
= Pi + x
Hence, x = P – Pi ........(ii)
Pressure of A, PA at time t is obtained by substitution of equation (ii) into the (i) equation. Thus
PA = Pi – (P – Pi)
= Pi – P + Pi
= 2Pi – P
The integrated rate law turns out to be
k = `2.303/t log_10 [A]_0/[A]_t`
The concentration is now expressed in terms of pressures.
Thus, [A]0 = Pi and [A]t = PA = 2Pi – P
Substitution gives in above
k = `2.303/t log_10 P_i/(2P_i - P)` .....(iii)
P is the total pressure of the reaction mixture at time t.
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