Graph between log k and 1 / T[k. is rate constant ( s -1) and T and temperature .( K )] is a straight line with OX =5, θ= tan -1(1 / 2.303). Hence -E a will be

Question

Graph between log k and 1 / T[k. is rate constant ( s -1) and T and temperature .( K )] is a straight line with OX =5, θ= tan -1(1 / 2.303). Hence -E a will be (A) 2.303 × 2 cal (B) 2 / 2.303 cal (C) 2 cal (D) None

Tardigrade Admin · Accepted Answer

log k= log A-(E a /2.303 R T)(y=c+m x) Slope =(-E a /2.303 R)=(1/2.303) (given) ( tan θ=(1/2.303)) -E a =2.303 R × Slope =2.303 × (R/2.303)=R=2 cal

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

$2.303 \times 2 c a l$

$2/2.303 c a l$

$2 c a l$

None

$lo g k = lo g A - \frac{E _{a}}{2.303 RT} (y = c + m x)$
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
$= 2.303 \times \frac{R}{2.303} = R = 2 c a l$

Q. Graph between logk and 1/T[k is rate constant (s−1) and T and temperature (K)] is a straight line with OX=5,θ=tan−1(1/2.303). Hence −Ea​ will be

2.303×2cal

2/2.303cal

2cal