Posted by Odesa on .
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 s1 and the concentration of AdBr, c, is in units of M (mole L1). 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 s1.

Chemistry 
MIT EDX 3.091,
Noted.

Chemistry 
amar chauhan,
arate=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 r1.63*(.03091)^1.67
=4.91*10^3 m/s 
Chemistry 
MIT EDX 3.091x,
Please do not answer this post. This question is a part of the MIT edx 3.091 solid state chemistry curriculum, and posting or using this information is in direct violation of the honor code. Thank you :)