Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with \(t_{\frac 12} = 3.00\ hours\). What fraction of sample of sucrose remains after \(8 \ hours\)?
For a first order reaction,
\(k = \frac {2.303}{t} log \ \frac {[R]_o}{[R]}\)
It is given that, \(t_{\frac 12} = 3.00\ hours\)
\(k = \frac {0.693}{t_{1/2}}\)
Therefore, \(k = \frac {0.693}{3} h^{-1}\)
\(k = 0.231\ h^{-1}\)
Then, \(0.231 \ h^{-1} = \frac {2.303}{t} log \ \frac {[R]_o}{[R]}\)
⇒ \(log\ \frac { [R]_o}{[R]} =\frac { 0.231 h^{-1} \times 8 h}{2.303}\)
⇒ \(\frac {[R]_o}{[R]}\) = \(antilog \ (0.8024)\)
⇒ \(\frac {[R]_o}{[R]}\) = \(6.3445\)
⇒ \(\frac {[R]_o}{[R]}\) = \(0.1576 \ (approx)\)
⇒ \(\frac {[R]_o}{[R]}\) = \(0.158\)
Hence, the fraction of sample of sucrose that remains after 8 hours is \(0.158\).
(i) Write any two differences between order and molecularity.
(ii) What do you mean by pseudo order reaction?
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.