A 1st order reaction is carried out with an initial concentration of 1

1.	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» 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%.