Dinitrogen pentoxide (N2O5) decomposes
by first-order kinetics with a rate constant of
0.0064 s−1
at a certain temperature. What is
the half life for the decomposition of N2O5 at
this temperature?
Answer in units of s.
i did: ln(.003135/x)=(5/8.314)((1/300)-(1/487))
i got .00311 but its wrong
we're cucked
To find the half-life of a first-order reaction, you need to use the formula:
t(1/2) = (0.693 / k)
Where:
- t(1/2) is the half-life,
- k is the rate constant of the reaction.
In this case, the rate constant (k) is given as 0.0064 s^-1. Now you can substitute this value into the formula:
t(1/2) = (0.693 / 0.0064)
= 108.28125 s
Therefore, the half-life for the decomposition of N2O5 at this temperature is approximately 108.28 seconds (s).