For the reaction: 2N205 (g) - 4N)2(g) the rate Law is:
change in [02]/change in time = k[N205]

At 300K, the half life is 2.50 * 1064 second and the activation energy is 103.3kJ/mol. At what temperature is the half life 400. second?

Question

For the reaction: 2N205 (g) -> 4N)2(g)  the rate Law is:
change in [02]/change in time = k[N205]

At 300K, the half life is 2.50 * 1064 second and the activation energy is 103.3kJ/mol. At what temperature is the half life 400. second?

anonymous · Answer

someone help

aaronq · Answer

i've never done a problem like this but i guess you can use the first order equation to find k, for both scenarios  (t1/2=300 s, 400 s)

$$t _{1/2}=\frac{ \ln2 }{ k }$$

then use the 2-point form of the Arrhenius equation

$$\ln \left( \frac{ k _{1} }{ k_{2} } \right)=\frac{ E _{a} }{ R }\left( \frac{ 1 }{ T _{2} }-\frac{ 1 }{ T _{1} } \right)$$

anonymous · Answer

could you please help me solve it. got the equation, its just plugging in the right numbers

aaronq · Answer

you want me to hand it in for you too? :P