Question

help me solve this chemistry's maths..​

Answer 1

Question :-

A first order reaction takes 40 minutes for 20% decomposition . Calculate half life period .

Answer :-

$\implies \boxed{ \sf{ \green{t_{ \frac{1}{2} }} = 124.19 \: minutes \: }}$

To Find :-

→ Half life period

Explanation :-

For First order reaction -

$\to \boxed{\sf{ t = \red{ \frac{2.303}{k} } \green{ \log( \frac{a}{a - x} ) }}} \\ \\ \because \sf{decomposition \: 20\% \: out \: of \: 100\%} \\ \\ \leadsto \sf{t = \red{ \frac{2.303}{k} } \green{ \log( \frac{100}{100 - 20} )}} \\ \\ \red{\leadsto }\sf{ t = \frac{2.303}{k} \log( \frac{10}{8} ) } \\ \because \sf{t = 40 \: min.} \\ \\ \leadsto \sf{40 = \orange{ \frac{2.303}{k} }( \log10 - \log8)} \\ \\ \leadsto \sf{ 40 = \red{\frac{2.303}{k}}(1 - 3 \log2)} \\ \\ \leadsto \sf{ k = \frac{2.303}{40} (1 - 3 \times 0.301)} \\ \\ \leadsto \sf{ k = \frac{2.303}{40} \times 0.97}$

$\leadsto\boxed{\sf{k=0.00558 \:{min}^{ - 1}}}$

Now we know that ; For half life period

$\implies \pink{ \boxed{\sf{ \green{t_{ \frac{1}{2} }} = \purple{ \frac{0.693}{k}}}}} \\ \\ \implies \sf{t_{ \frac{1}{2} } } = \frac{0.693}{0.00558} \\ \\ \implies \boxed{ \sf{ \green{t_{ \frac{1}{2} }} = 124.19 \: minutes \: }}$

Answer 2

Answer:

Question :-

A first order reaction takes 40 minutes for 20% decomposition . Calculate half life period .

Answer :-

$\implies \boxed{ \sf{ \green{t_{ \frac{1}{2} }} = 124.19 \: minutes \: }}$

To Find :-

→ Half life period

Explanation :-

For First order reaction -

$\to \boxed{\sf{ t = \red{ \frac{2.303}{k} } \green{ \log( \frac{a}{a - x} ) }}} \\ \\ \because \sf{decomposition \: 20\% \: out \: of \: 100\%} \\ \\ \leadsto \sf{t = \red{ \frac{2.303}{k} } \green{ \log( \frac{100}{100 - 20} )}} \\ \\ \red{\leadsto }\sf{ t = \frac{2.303}{k} \log( \frac{10}{8} ) } \\ \because \sf{t = 40 \: min.} \\ \\ \leadsto \sf{40 = \orange{ \frac{2.303}{k} }( \log10 - \log8)} \\ \\ \leadsto \sf{ 40 = \red{\frac{2.303}{k}}(1 - 3 \log2)} \\ \\ \leadsto \sf{ k = \frac{2.303}{40} (1 - 3 \times 0.301)} \\ \\ \leadsto \sf{ k = \frac{2.303}{40} \times 0.97}$

$\leadsto\boxed{\sf{k=0.00558 \:{min}^{ - 1}}}$

Now we know that ; For half life period

$\implies \pink{ \boxed{\sf{ \green{t_{ \frac{1}{2} }} = \purple{ \frac{0.693}{k}}}}} \\ \\ \implies \sf{t_{ \frac{1}{2} } } = \frac{0.693}{0.00558} \\ \\ \implies \boxed{ \sf{ \green{t_{ \frac{1}{2} }} = 124.19 \: minutes \: }}$