Question

$\text{\large{\red{\underline{Question:-}}}}$

Consider a certain reaction A→ Products with,
$k \: = 2.0 \times {10}^{ - 2} \: {s}^{ - 1}$
Calculate the concentration of A remaining after 100 s, if the initial concentration of A is
${1.0 \: mol \: L}^{ - 1}$
Don't post irrelevant answers please !!​

Answer 1

$\bf{\huge{\red{\underline{Answer:-}}}}$

[A] = 0.135 M

$\text{\large{\orange{\underline{Given:-}}}}$

$k \: = 2.0 \: \times {10}^{ - 2} \: {s}^{ - 1}$

$t \: = 100s$

$[A] 0 \: = 1.0 \: mol \: {L}^{ - 1} = 1.0M\:$

$\text{\large{\orange{\underline{Solution:-}}}}$

$\red{\underline{For\:the\:first\:order\:reaction,}}$

$t \: = \frac{2.303}{k} log( \frac{[A]0}{[A]} ) \: \: or \: \: log( \frac{[A]0}{[A] } ) \: = \frac{k \times t}{2.303}$

$\red{log(\frac{[A]0}{[A]})}$ = ${\red{\frac{(2.0\times{10}^{-2}{s}^{-1})(100\:s)}{2.303}}}$ = $\red{\underline{0.8684}}$

$\red{\frac{[A]0}{[A]}}$ = Antilog 0.8684 = $\red{\underline{7.3858}}$

[A] = ${\frac{[A0]}{7.3858}}$ = ${\frac{[1M]}{7.3858}}$ = 0.135 M

$\red{\underline\therefore{[A]=0.135\:M}}$

Hence, the concentration of A remaining after 100 seconds is 0.135 M.

Answer 2

$\bf{\huge{\red{\underline{Answer:-}}}}$

[A] = 0.135 M

$\text{\large{\orange{\underline{Given:-}}}}$

$k \: = 2.0 \: \times {10}^{ - 2} \: {s}^{ - 1}$

$t \: = 100s$

$[A] 0 \: = 1.0 \: mol \: {L}^{ - 1} = 1.0M\:$

$\text{\large{\orange{\underline{Solution:-}}}}$

$\red{\underline{For\:the\:first\:order\:reaction,}}$

$t \: = \frac{2.303}{k} log( \frac{[A]0}{[A]} ) \: \: or \: \: log( \frac{[A]0}{[A] } ) \: = \frac{k \times t}{2.303}$

$\red{log(\frac{[A]0}{[A]})}$ = ${\red{\frac{(2.0\times{10}^{-2}{s}^{-1})(100\:s)}{2.303}}}$ = $\red{\underline{0.8684}}$

$\red{\frac{[A]0}{[A]}}$ = Antilog 0.8684 = $\red{\underline{7.3858}}$

[A] = ${\frac{[A0]}{7.3858}}$ = ${\frac{[1M]}{7.3858}}$ = 0.135 M

$\red{\underline\therefore{[A]=0.135\:M}}$

Hence, the concentration of A remaining after 100 seconds is 0.135 M.