Solution Found!
The conversion of ammonium cyanide to urea is a
Chapter 3, Problem 10E(choose chapter or problem)
The conversion of ammonium cyanide to urea is a second-order reaction. This means that the concentration C of ammonium cyanide at time t is given by \(1 / C=k t+1 / C_{0}\), where \(C_{0}\) is the initial concentration and k is the rate constant. Assume the initial concentration is known to be 0.1 mol/L exactly. Assume that time can be measured with negligible uncertainty.
a. After 45 minutes, the concentration of ammonium cyanide is measured to be \(0.0811 \pm 0.0005 \mathrm{~mol/L}\). Estimate the rate constant k, and find the uncertainty in the estimate.
b. Use the result in part (a) to estimate the time when the concentration of ammonium cyanide will be 0.0750 mol/L, and find the uncertainty in this estimate.
Equation Transcription:
Text Transcription:
1/C=kt+1/C_0
C_0
0.1 mol/L
0.0811{+/-}0.0005 mol/L
0.0750 mol/L
Questions & Answers
QUESTION:
The conversion of ammonium cyanide to urea is a second-order reaction. This means that the concentration C of ammonium cyanide at time t is given by \(1 / C=k t+1 / C_{0}\), where \(C_{0}\) is the initial concentration and k is the rate constant. Assume the initial concentration is known to be 0.1 mol/L exactly. Assume that time can be measured with negligible uncertainty.
a. After 45 minutes, the concentration of ammonium cyanide is measured to be \(0.0811 \pm 0.0005 \mathrm{~mol/L}\). Estimate the rate constant k, and find the uncertainty in the estimate.
b. Use the result in part (a) to estimate the time when the concentration of ammonium cyanide will be 0.0750 mol/L, and find the uncertainty in this estimate.
Equation Transcription:
Text Transcription:
1/C=kt+1/C_0
C_0
0.1 mol/L
0.0811{+/-}0.0005 mol/L
0.0750 mol/L
ANSWER:
Step 1 of 3:
The experiment under consideration here is conversion of ammonium cyanide to urea.
The concentration C of ammonium cyanide at time t is given by,
=kt+
where is the initial concentration of ammonium cyanide, k is rate constant.
It is also given that =0.1mol/L.