2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-1998

  • question_answer
    After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is\[\text{1}\text{.155}\times {{10}^{-3}}{{\sec }^{-1}}\]?

    A)  600                                       

    B)  100

    C)  60                                         

    D)  10

    Correct Answer: A

    Solution :

                     Rate constant \[k=1.155\times {{10}^{-3}}{{\sec }^{-1}}\] \[k=\frac{2.303}{t}\log \frac{a}{(a-x)}\]\[\because \]\[a=a,(a-x)=\frac{a}{2}\]                 \[{{t}_{1/2}}=\frac{2.303}{k}\log \frac{a}{a/2}\] \[=\frac{2.303}{1.155\times {{10}^{-3}}}\log 2\] \[=\frac{2.303}{1.155\times {{10}^{-3}}}\times 0.3010\] \[=\frac{0.693\times {{10}^{3}}}{1.155}\] \[=0.6\times {{10}^{3}}=600\sec \]                 or            \[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{1.155\times {{10}^{-3}}}\]                                 \[=600\,\sec \]


You need to login to perform this action.
You will be redirected in 3 sec spinner