Graph between log k and 1 T [ k is rate constant ( s - 1 ) and T the temperature (K)] is a straight line. OX = 5, = tan - 1 ( 1 2 . 303 ) . Hence - E a will be

Question

Graph between log k and $\frac{1}{T}$ [ $k$ is rate constant $(s^{- 1})$ and $T$ the temperature (K)] is a straight line. OX = 5, $θ = \tan^{- 1} (\frac{1}{2.303})$ . Hence $- E_{a}$ will be

Smart Achievers · Accepted Answer

(3)

$\log k = \log A - \frac{E_{a}}{2.303 R T} (y = c + m x)$

$Slope = \frac{- E_{a}}{2.303 R} = \frac{1}{2.303} (given) (\tan θ = \frac{1}{2.303})$

$- E_{a} = 2.303 R \times slope$

$= 2.303 \times \frac{R}{2.303} = R = 2 cal$

Solution

Q.
Graph between log k and $\frac{1}{T}$ [ $k$ is rate constant $(s^{- 1})$ and $T$ the temperature (K)] is a straight line. OX = 5, $θ = \tan^{- 1} (\frac{1}{2.303})$ . Hence $- E_{a}$ will be

1 $2.303 \times 2 cal$

2 $\frac{2}{2.303 cal}$

3 $2 cal$

4 $\frac{2.303}{2} cal$

Ans.
(3)

$\log k = \log A - \frac{E_{a}}{2.303 R T} (y = c + m x)$

$Slope = \frac{- E_{a}}{2.303 R} = \frac{1}{2.303} (given) (\tan θ = \frac{1}{2.303})$

$- E_{a} = 2.303 R \times slope$

$= 2.303 \times \frac{R}{2.303} = R = 2 cal$

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Solution

Q. Graph between log k and 1T [k is rate constant (s-1) and T the temperature (K)] is a straight line. OX = 5, θ=tan-1(12.303). Hence -Ea will be

1 2.303×2 cal

2 22.303 cal

3 2 cal

4 2.3032 cal

Ans. (3) logk=logA-Ea2.303RT (y=c+mx) Slope=-Ea2.303R=12.303 (given) (tanθ=12.303) -Ea=2.303R×slope =2.303×R2.303=R=2 cal