Administratium bromide (AdBr) decomposes readily at temperatures exceeding 37°C. The figure below shows how the rate of reaction varies with the concentration of AdBr. The rate, r, is in units of M s-1 and the concentration of AdBr, c, is in units of M (mole L-1). The slope has a value of a 1.67 and the intercept has a value of 0.490.
(a) What is the order of reaction?
(b) Calculate the instant rate of decomposition of AdBr when its concentration is 0.03091 M. Express your answer in M s-1.
3 answers
Noted.
a--rate=k(AdBr)^r
ln (rate)=r ln ((AdBr)+ln k
r=1.67
ln k=0.490=>k=1.63
order of reaction=1.63
b-- and rate of decomposition r-1.63*(.03091)^1.67
=4.91*10^-3 m/s
ln (rate)=r ln ((AdBr)+ln k
r=1.67
ln k=0.490=>k=1.63
order of reaction=1.63
b-- and rate of decomposition r-1.63*(.03091)^1.67
=4.91*10^-3 m/s
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