Slope of straight line in figure- 1 is − 1 × 1 0 − 1 sec − 1 . What is the half-life of the reaction? (ℓn 2 = 0.693)

Kt = ln A 0 − ln A t ln A t = ln A 0 − Kt slope = − K = − ( 1 × 1 0 − 1 ) t 2 1 = 1 0 − 1 0.693 = 6.93 sec

Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsFirst Order ReactionsMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

Read the following passage carefully and answer the questions given after the passage :

Following reaction takes place via first order kinetics

$2 N_{2} O_{5} ⟶ 4 N O_{2} (g) + O_{2} (g)$

0.2 mole of $N_{2} O_{5}$ was taken in a vessel of capacity 1 L and heating to a certain temperature of T Kelvin at which the reaction begins. The concentration of $N_{2} O_{5}$ is measured at different intervals of time by withdrawing samples of reaction mixture. A plot of $ℓ n [N_{2} O_{5}] v / s$ time is shown is figure-1. The value of rate constant ( k ) of the reaction is determined from the slope of the linear plot. The same experiment is repeated at five different temperature and the value of the rate constants are calculated at each temperature. A plot of $ℓ nk / s$ (temperature) $^{- 1}$ is shown in figure-2 :

Slope of straight line in figure- 1 is $- 1 \times 1 0^{- 1} sec^{- 1}$ . What is the half-life of the reaction?

(ℓn 2 = 0.693)

Answer:

6.93

Solution:

$Kt = ln A_{0} - ln A_{t}$

$ln A_{t} = ln A_{0} - Kt$

slope $= - K = - (1 \times 1 0^{- 1})$

$t_{\frac{1}{2}} = \frac{0.693}{1 0 ^{- 1}} = 6.93 sec$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:First Order Reactions

⏱ 2mℹ️ Source: QB

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