Ammonium cyanate, NH4NCO, rearranges in water to form urea, (NH2)2CO, through a

Question

Ammonium cyanate, NH4NCO, rearranges in water to form urea, (NH2)2CO, through a second-order process. The rate constant for the process is 0.0113 M-1min-1. If the original concentration of NH4NCO is 0.229 M, how much remain after 150 minutes? Ammonium cyanate, NH4NCO, rearranges in water to form urea, (NH2)2CO, through a second-order process. The rate constant for the process is 0.0113 M-1min-1. If the original concentration of NH4NCO is 0.229 M, how much remain after 150 minutes? Ammonium cyanate, NH4NCO, rearranges in water to form urea, (NH2)2CO, through a second-order process. The rate constant for the process is 0.0113 M-1min-1. If the original concentration of NH4NCO is 0.229 M, how much remain after 150 minutes?

Explanation / Answer

rate constant= K= 0.0113 M-1min-1

initial concentration of NH4NCO = N0 = 0.229M

time = 150 minutes

for second order reaction

half life reaction t1/2=1/k [A]0

t1/2= 1/0.0113x0.229= 386.44 min

total life period = 150 minutes

number of half lives = total life period/half life period

number of half lives = n= 150/386.44 = 0.388

N= N0/2^n = 0.229/2^0.388 = 0.229/ 1.309 =0.175

remainign concentration of NH4NCO = 0.175 M

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