
- A$4$
- B$5$
- ✓$6$
- D$7$


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${H_2}O\left( g \right) + C\left( s \right) \to CO\left( g \right) + {H_2}\left( g \right);\Delta H = 131\,kJ$
$CO\left( g \right) + \frac{1}{2}{O_2}\left( s \right) \to C{O_2}\left( g \right);\Delta H = - 282\,kJ$
${H_2}\left( g \right) + \frac{1}{2}{O_2}\left( s \right) \to {H_2}O\left( g \right);\Delta H = - 242\,kJ$
$C(s) + O_2(g) \to CO_2(g); \Delta H = X\,kJ$
The value of $X$ will be......$kJ$
(at. mass $Ag =108; Br = 80$)
Given that the lonic product of $Ni ( OH )_{2}$ is $2 \times 10^{-15}$
${C_6}{H_6}(l)\, + \,\frac{{15}}{2}\,{O_2}(g)\, \to \,6\,C{O_2}(g)\, + \,3{H_2}O\,(g)$
Signs of $\Delta H,\,\Delta S$ and $\Delta G$ for the above reaction will be