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# $H_2O(g) + C(s) \rightarrow CO(g) + H_2(g)$ ; $\Delta H = 131$ kJ. $CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -282$ kJ. $H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g)$ ; $\Delta H = -242$ kJ. $C(s) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = X$ kJ. Based on the above thermochemical equations find the value of X.

(a) +393 kJ
(b) -655 kJ
(c) -393 kJ
(d) +655 kJ

$H_2O(g) + C(s) \rightarrow CO(g) + H_2(g)$ ; $\Delta H = 131$ kJ.
$CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -282$ kJ.
$H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g)$ ; $\Delta H = -242$ kJ.
To obtain this equation, $C(s) + O_2(g) \rightarrow CO_2(g)$, the above equations must be added,
Thus, we need to add the enthalpies of the three equations in order to obtain the enthalpy of the fourth equation.
$\Delta H = X kJ = 131 + (-282) + (-242) = -393$ kJ
$\therefore C(s) + O_2(g) \rightarrow CO_2(g)$ ; $\Delta H = -393$ kJ.