Question

The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be at 298oK? What will the half-life be at both temperatures?

Answer 1

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature $T_1 =$ 1001 K

Final temperature $T_2 =$ 298 K

Applying the equation of Arrhenius theory.

$In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})$

where ;

R gas constant = 8.314  J/K/mol

$In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})$

$In \dfrac{k_2}{0.380 }= -70.8655$

$\dfrac{k_2}{0.380 }= e^{-70.8655}$

$\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}$

${k_2}= 1.67303256 \times 10^{-31} \times {0.380 }$

${k_2}= 6.3575 \times 10^{-32}$  /M .sec

Half life:

At 1001 K.

$t_{1/2} = \dfrac{In_2}{k_1}$

$t_{1/2} = \dfrac{0.693}{0.38}$

$t_{1/2} =$ 1.82368 secc

At 298 K:

$t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}$

$t_{1/2} =1.09 \times 10^{31} \ sec$