Question

A 1st order reaction is carried out with an initial concentration of 10 mol per liter and 80%of the reactant is changes into product .now if the same reaction is carried ot wih an initial concentration of 5 mol per liter for the same period the percentage of the reactant changing into product is?

Answer 1

Solution:

Order of the reaction = 1

Case - 1

$\sf{\implies k = \dfrac{2.303}{t} \log_{10} \Bigg(\dfrac{100}{100-80}\Bigg)}$

$\sf{\implies k = \dfrac{2.303}{t} \log_{10}\Bigg(\dfrac{10}{2}\Bigg)}$

$\sf{\implies \dfrac{2.303}{t} \times 0.70\;\;\;\;.........(1)}$

Case - 2

$\sf{\implies k = \dfrac{2.303}{t} \log_{10} \Bigg(\dfrac{5}{x}\Bigg)}$

$\sf{\implies k = \dfrac{2.303}{t} \Big[\log_{10}5-\log_{10}x\Big]}$

$\sf{\implies \dfrac{2.303}{t}\Big[0.70-\log_{10}x\Big]\;\;\;\;........(2)}$

Dividing Eq (1) by Eq (2),

$\sf{\implies \dfrac{k}{k}=\dfrac{\dfrac{2.303 \times [0.70]}{t}}{\dfrac{2.303 \times [0.70-\log_{10}x]}{t}}}$

$\sf{\implies 1 = \dfrac{0.70}{0.70-\log_{10}x}}$

On solving it we get,

$\sf{\implies \log_{10}x = 0}$

$\sf{\implies x = Anti \log 0} \\ \\ \sf{\implies x = 1}$

Now, keeping x = 1 in eq(2), we analyse that Eq (1) = Eq (2). Hence % of reactant changing to product will be same i.e 80%.

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Order\:of\:reaction=1}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\bold{First\:case:}\\\implies k = \frac{2.303}{t} log_{10}( \frac{100}{100 - 8} ) \\ \\ \implies k = \frac{2.303}{t} log_{10}( \frac{10}{2} ) \\ \\ \implies k = \frac{2.303}{t} \times 0.7 - - - - - (1) \\ \\ \bold{Second\:case:} \\ \implies k = \frac{2.303}{t} log_{10}(\frac{5}{x} ) \\ \\ \implies k = \frac{2.303}{t} (log_{10}5 - log_{10}x) \\ \\ \implies k = \frac{2.303}{t} (0.7 - log_{10}x) - - - - - (2) \\ \\ \bold{dividing \: (1) \: and \: (2)} \\ \\ \implies \frac{ \cancel k}{ \cancel k} = \frac{ \frac{2.303 \times (0.7)}{ \cancel t} }{ \frac{2.303 \times (0.7 - log_{10}x)}{ \cancel t} } \\ \\ \implies 1 = \frac{0.7}{0.7 - log_{10}x} \\ \\ \implies log_{10}x = 0 \\ \\ \implies x = {10}^{0} \\ \\ \bold{\implies x = 1}$