Given :

$$ \begin{aligned} & \left.\Delta \mathrm{H}^{\ominus}{ }_{\text {sub }}[\mathrm{C} \text { (graphite })\right]=710 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\ominus}=436 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}}=\mathrm{C} \mathrm{H}^{\ominus}=611 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$

The $\Delta \mathrm{H}_{\mathrm{f}} \ominus$ for $\mathrm{CH}_2=\mathrm{CH}_2$ is_________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)

$$ \text { If the domain of the function } f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right) \text { is }[\alpha, \beta) \text {, then } \alpha^2+4 \beta \text { is equal to } $$

Let $a_1, a_2, a_3, \ldots$. be a G.P. of increasing positive numbers. If $a_3 a_5=729$ and $a_2+a_4=\frac{111}{4}$, then $24\left(a_1+a_2+a_3\right)$ is equal to

The sum $1+3+11+25+45+71+\ldots$ upto 20 terms, is equal to