In a closed flask at 600 K , one mole of $\mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ attains equilibrium as given below :

$$ \mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}_2(\mathrm{~g}) $$

At equilibrium, $75 \% \mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ was dissociated and the total pressure is 1 atm . The magnitude of $\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) at this temperature is $\_\_\_\_$ . (Nearest Integer)

(Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ln 10=2.3, \log 2=0.3, \log 3=0.48, \log 5=0.69, \log 7=0.84$ )

Decomposition of a hydrocarbon follows the equation $\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}$. The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)

Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x)=\frac{2 x^2-3 x+2}{3 x^2+x+3}$. Then $f$ is :

Consider the quadratic equation $\left(n^2-2 n+2\right) x^2-3 x+\left(n^2-2 n+2\right)^2=0, n \in \mathbf{R}$. Let $\alpha$ be the minimum value of the product of its roots and $\beta$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $\alpha$ and the common ratio is $\frac{\alpha}{\beta}$, is :