Question

Sucrose decomposes to fructose and glucose in acid solution. When In (sucrose) is plotted vs. time, a straight line with slope of -0 208 hr results. What is the rate law for the reaction? Rate = 0.208 hr(sucrose) ob Rate = 0.208 hr [sucrose) Woc Rate = 0,0433 hr (sucrose d Rate = 0.0433 hr (sucrose] O Rate = 0.208 mol L-Throucrosojº

Answer 1

integrated rate law for 2nd order reaction:

1/[A] = 1/[A]o + k*t

so, for 2nd order 1/concentration vs time is linear

integrated rate law for 1st order reaction:

ln [A] = –kt + ln [A]_o

so, for 1st order ln(concentration) vs time is linear

integrated rate law for zero order reaction:

[A] = [A]o – k*t

so, for 0 order concentration vs time is linear

We have ln[sucrose] vs t as constant.

So, this is 1st order reaction.

Slope = -k

-0.208 hr-1 = -k

k = 0.208 hr-1

Rate law is:

Rate = k [sucrose]

Rate = 0.208 hr-1 [sucrose]

Answer: b