Answer:
0.834 M.
Explanation:
Hello!
In this case, since this is a first-order reaction, we can infer that the half-life and the rate constant are related via:
[tex]t_{1/2}=\frac{ln(2)}{k}[/tex]
Thus, we are able to compute the rate constant:
[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{32s}\\\\k=0.0217s^{-1}[/tex]
Next, since we have the resulting concentration of the reactant and we need the initial one, we proceed as shown below with the rate law:
[tex]A=A_0exp(-kt)\\\\A_0=\frac{A}{exp(-kt)} \\\\A_0=\frac{0.062M}{exp(-0.0217s^{-1}*2.0min*\frac{60s}{1min} )}\\\\A_0=0.834M[/tex]
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