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Standard XII

Chemistry

Arrhenius Equation

On introducin...

Question

On introducing a catalyst at $500$ $K$ , the rate of a first order reaction increases by $1.718$ times. The activation energy in the presence of a catalyst is $6.05$ $k J$ ${m o l}^{- 1}$ . The slope of the plot of $ln k ({s e c}^{- 1})$ against $1 / T$ in the absence of catalyst is:

+ 1

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- 1

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+ 1000

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- 1000

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Solution

The correct option is D $- 1000$
$\frac{R a t e i n p r e s e n c e o f c a t a l y s t}{R a t e i n a b s c e n c e o f c a t a l y s t} = A n t i l o g [\frac{+ Δ E}{2.303 R T}]$
$1.718 = A n t i l o g \frac{E_{a} - E_{p}}{2.303 \times 8.314 \times 500}$
$E_{a} - E_{p} = 2.25$ $k J$
$E_{a} = E_{p} + 2.25 = 6.05 + 2.25 = 8.30$ $k J$ ${m o l}^{- 1}$
$= 8.3$ $k J$ ${m o l}^{- 1}$
$ln k = ln A - \frac{E_{a}}{R} \times \frac{1}{T}$
Slope $= \frac{- E_{a}}{R} = \frac{- 8.3 \times 1000}{8.3} = - 1000$

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Similar questions

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On introducing a catalyst at 500 K, the rate of a first order reaction increases by 1.718 times. The activation energy in the presence of a catalyst is 6.05 kJ mol−1. The slope of the plot of lnk(sec−1) against 1/T in the absence of catalyst is:

The correct option is D −1000RateinpresenceofcatalystRateinabscenceofcatalyst=Antilog[+ΔE2.303RT]1.718=AntilogEa−Ep2.303×8.314×500Ea−Ep=2.25 kJEa=Ep+2.25=6.05+2.25=8.30 kJ mol−1 =8.3 kJ mol−1lnk=lnA−EaR×1TSlope =−EaR=−8.3×10008.3=−1000

On introducing a catalyst at $500$ $K$ , the rate of a first order reaction increases by $1.718$ times. The activation energy in the presence of a catalyst is $6.05$ $k J$ ${m o l}^{- 1}$ . The slope of the plot of $ln k ({s e c}^{- 1})$ against $1 / T$ in the absence of catalyst is: