Question

reacting with water in an acidic solution at a particular temperature, compound a decomposes into compounds b and c according to the law of uninhibited decay. an initial amount of 0.60 m of compound a decomposes to 0.57 m in 30 minutes. how much of compound a will remain after 3 hours? how long will it take until 0.10 m of compound a remains? after 3 hours, the amount of compound a remaining will be \\(\square\\) m. (do not round until the final answer. then round to the nearest hundredth as needed.)

Explanation:

Step1: Find the decay constant \(k\)

The formula for uninhibited decay is \(A = A_0e^{-kt}\). We know that \(A_0 = 0.60\space M\), \(A = 0.57\space M\), and \(t = 30\space min=\frac{30}{60}=0.5\space h\).
Substitute into the formula: \(0.57=0.60e^{-k\times0.5}\).
Divide both sides by \(0.60\): \(\frac{0.57}{0.60}=e^{- 0.5k}\).
Take the natural logarithm of both sides: \(\ln(\frac{0.57}{0.60})=-0.5k\).
Solve for \(k\): \(k =-\frac{\ln(\frac{0.57}{0.60})}{0.5}\).
\(\ln(\frac{0.57}{0.60})=\ln(0.95)\approx - 0.05129\), so \(k=\frac{0.05129}{0.5}=0.10258\space h^{-1}\).

Step2: Find the amount after \(t = 3\space h\)

Use the formula \(A = A_0e^{-kt}\) with \(A_0 = 0.60\space M\), \(k = 0.10258\space h^{-1}\), and \(t = 3\space h\).
\(A=0.60e^{-0.10258\times3}\).
\(A = 0.60e^{-0.30774}\).
Since \(e^{-0.30774}\approx0.735\), then \(A = 0.60\times0.735 = 0.441\space M\).

Step3: Find the time \(t\) when \(A = 0.10\space M\)

Use the formula \(A = A_0e^{-kt}\), substitute \(A = 0.10\space M\), \(A_0 = 0.60\space M\), and \(k = 0.10258\space h^{-1}\).
\(0.10=0.60e^{-0.10258t}\).
Divide both sides by \(0.60\): \(\frac{0.10}{0.60}=e^{-0.10258t}\).
Take the natural logarithm of both sides: \(\ln(\frac{0.10}{0.60})=-0.10258t\).
\(\ln(\frac{1}{6})\approx - 1.7918\), so \(t=\frac{-1.7918}{- 0.10258}\approx17.5\space h\).

Answer:

After 3 hours, the amount of compound A remaining is \(0.44\space M\). It will take approximately \(17.5\space h\) until \(0.10\space M\) of compound A remains.