13) The reaction CH_(3)-Nequiv Carrow CH_(3)-Cequiv N is a first-order reaction . At 2303^circ C.k=6.29ast 10^-4s-1 If [CH_(3)-N=C] is 1.00times 10^-3 initially. [CH_(3)-N=Cl __ Mafter 1.000times 10^3s A) 5.33times 10-4 B) 2.34times 10-4 C) 1.88times 10^-3 D) 4.27times 10.3

To solve this problem, we can use the integrated rate law for a first-order reaction:<br /><br />\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]<br /><br />where:<br />- \([A]_0\) is the initial concentration of the reactant,<br />- \([A]\) is the concentration of the reactant at time \(t\),<br />- \(k\) is the rate constant,<br />- \(t\) is the time.<br /><br />Given:<br />- \([CH_3-N \equiv C]_0 = 1.00 \times 10^{-3} \, \text{M}\)<br />- \(k = 6.29 \times 10^{-4} \, \text{s}^{-1}\)<br />- \(t = 1.000 \times 10^3 \, \text{s}\)<br /><br />We need to find \([CH_3-C \equiv N]\) after \(1.000 \times 10^3 \, \text{s}\).<br /><br />First, calculate the concentration of the reactant remaining after \(1.000 \times 10^3 \, \text{s}\):<br /><br />\[ \ln \left( \frac{[CH_3-N \equiv C]}{[CH_3-N \equiv C]_0} \right) = -kt \]<br /><br />\[ \ln \left( \frac{[CH_3-N \equiv C]}{1.00 \times 10^{-3}} \right) = -(6.29 \times 10^{-4}) \times (1.000 \times 10^3) \]<br /><br />\[ \ln \left( \frac{[CH_3-N \equiv C]}{1.00 \times 10^{-3}} \right) = -0.629 \]<br /><br />\[ \frac{[CH_3-N \equiv C]}{1.00 \times 10^{-3}} = e^{-0.629} \]<br /><br />\[ [CH_3-N \equiv C] = 1.00 \times 10^{-3} \times e^{-0.629} \]<br /><br />\[ [CH_3-N \equiv C] \approx 1.00 \times 10^{-3} \times 0.527 \]<br /><br />\[ [CH_3-N \equiv C] \approx 5.27 \times 10^{-4} \, \text{M} \]<br /><br />Now, calculate the concentration of \([CH_3-C \equiv N]\):<br /><br />\[ [CH_3-C \equiv N] = [CH_3-N \equiv C]_0 - [CH_3-N \equiv C] \]<br /><br />\[ [CH_3-C \equiv N] = 1.00 \times 10^{-3} - 5.27 \times 10^{-4} \]<br /><br />\[ [CH_3-C \equiv N] = 4.73 \times 10^{-4} \, \text{M} \]<br /><br />Therefore, the correct answer is:<br /><br />A) \(5.33 \times 10^{-4}\)