Question

Chem 1212K-D hemistryCourse Home CHW 5 Chapter 14 Question 32-Algorithmic Constants Part A Aparticular first-order reaction has a rate oonstant of 1.35 × 10,-1 at 25.0 °C.What is the magnitude of k at 950°CifEa-555 kJ/mol. 576 O 285 x 104 O 433 x 1087 136 x 102 O 958 x 103 Submit Request Answer Provide Feedback

Answer 1

Rate constant k₁=1.35x10^-2 s

Temperature=T₁=25^oC=25+273 K=298 K

Rate constant k₂=?

Temperature T₂=95^oC=95+273 K=368 K

Activation energy E_a=55.5 kJ/mol=55500 J/mol (1 kJ=1000 J)

We know from arrhenious equation

where k=Rate constant at temperature T

A=Frequency factor

R=Gas constant=8.314 J/Kmol

T=Temperature in K

At Temperatures T₁ and T₂

$k_1=Ae^{\frac{-Ea}{RT_1}}$ and $k_2=Ae^{\frac{-Ea}{RT_2}}$

Dividing k₂ by k₁

_{$\frac{k_2}{k_1}=\frac {e^{\frac{-Ea}{RT_2}}}{e^{\frac{-Ea}{RT_1}}}$}

Taking log_e on both sides we get

$log_e \frac{k_2}{k_1}=\frac {-Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1})$ ...equation (1)

Putting the given values in equation (1)

$log_e \frac{k_2}{1.35\times10^2s^{-1}}=\frac {-55500 J}{8.314 JK^{-1}mol^{-1}}(\frac{1}{368 K}-\frac{1}{298 K})$

$log_e k_2 - log_e1.35\times10^2s^{-1}=\frac {-55500 Jmol^{-1}}{8.314 JK^{-1}mol^{-1}}(\frac{1}{368 K}-\frac{1}{298 K})$

log_ek₂ - 4.91= $(-6675.49 K) (\frac{298-368 }{368\times298 K})$

log_ek₂ - 4.91= $(-6675.49 K)(\frac{-70}{109664K})$ =4.26

log_ek₂=4.26+4.91=9.17

k₂=e^9.17=9604.62 s^-1

which is closest to the last option.

So the last option is correct.