Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3 \text{ hours}$. The percentage of sucrose remaining after $6 \text{ hours}$ is . . . . . . . (Nearest integer) (Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$)

The half-life period $t_{1/2}$ for a first-order reaction is given by:

$A$ reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \ min^{-1}$. If we start with $[A] = 0.5 \ mol \ L^{-1}$,when would $[A]$ reach the value $0.05 \ mol \ L^{-1}$? (in $min$)

The initial concentration of $N_2O_5$ in the following first order reaction $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + 1/2O_{2(g)}$ was $1.24 \times 10^{-2} \, mol \, L^{-1}$ at $318 \, K$. The concentration of $N_2O_5$ after $60 \, minutes$ was $0.20 \times 10^{-2} \, mol \, L^{-1}$. Calculate the rate constant of the reaction at $318 \, K$.

For a first-order reaction,the time taken for the concentration of the reactant to reach $3/4$ of its initial value is $t_{1/4}$. If the rate constant for the reaction is $K$,then $t_{1/4}$ can be expressed as: (in $/K$)

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)