Figure shows a graph in log10 K vs (1/T) where K is rate constant and T is temperature. The straight line BC has slope, |tanθ|=(1/2.303) and an intercept of 5 on y-axis. Thus EA, the energy of activation is:

Question

Figure shows a graph in log10 K vs (1/T) where K is rate constant and T is temperature. The straight line BC has slope, |tanθ|=(1/2.303) and an intercept of 5 on y-axis. Thus EA, the energy of activation is: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/96c3c7e22e00d52ec95d26576e4270e2-.png" /> (A) 2.303 × 2 cal (B) 2/2.303 cal (C) 2 cal (D) None of these

Tardigrade Admin · Accepted Answer

log10K vs.(1/T) is linear; slope=(-EA/2.303R) intercept = log10 A Since, |tan θ|=(1/2.303) ∴ (EA/2.303R)=(1 /2.303) ⇒ Ea=R=2 cal

Q. Figure shows a graph in $l o g_{10} K v s \frac{1}{T}$ where $K$ is rate constant and $T$ is temperature. The straight line $BC$ has slope, $∣ t an θ ∣ = \frac{1}{2.303}$ and an intercept of $5$ on y-axis. Thus $E_{A}$ , the energy of activation is:

$2.303 \times 2 c a l$

$2/2.303 c a l$

$2 c a l$

None of these

$l o g_{10} K v s . \frac{1}{T}$ is linear;
slope $= \frac{- E _{A}}{2.303 R}$ intercept $= l o g_{10} A$
Since, $∣ t an θ ∣ = \frac{1}{2.303}$
$∴ \frac{E _{A}}{2.303 R} = \frac{1}{2.303} \Rightarrow E_{a} = R = 2 c a l$

Q. Figure shows a graph in log10​KvsT1​ where K is rate constant and T is temperature. The straight line BC has slope, ∣tanθ∣=2.3031​ and an intercept of 5 on y-axis. Thus EA​, the energy of activation is:

2.303×2cal

2/2.303cal

2cal