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(D) Given:$(i) M + \frac{1}{2} O_2 	o MO, \Delta H_1 = -351.4 \ kJ$$(ii) X + \frac{1}{2} O_2 	o XO, \Delta H_2 = -90.8 \ kJ$To find the heat of reac

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$260.6$

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(D) Given:$(i) M + \frac{1}{2} O_2 	o MO, \Delta H_1 = -351.4 \ kJ$$(ii) X + \frac{1}{2} O_2 	o XO, \Delta H_2 = -90.8 \ kJ$To find the heat of reaction for $M + XO 	o MO + X$,we subtract equation $(ii)$ from equation $(i)$:$(M + \frac{1}{2} O_2) - (X + \frac{1}{2} O_2) 	o MO - XO$$M - X 	o MO - XO$$M + XO 	o MO + X$The enthalpy change $\Delta H = \Delta H_1 - \Delta H_2$$\Delta H = (-351.4) - (-90.8) = -351.4 + 90.8 = -260.6 \ kJ$Since the reaction is exothermic,the heat released is $260.6 \ kJ$.

Substance	$H_{2(g)}$	$C_{\text{(graphite)}}$	$C_{2}H_{6(g)}$
$\Delta_{c}H^{\Theta} / (kJ \ mol^{-1})$	$-286.0$	$-394.0$	$-1560.0$

The following thermochemical reactions are given:
$M + \frac{1}{2} O_2 \to MO + 351.4 \ kJ$ $(i)$
$X + \frac{1}{2} O_2 \to XO + 90.8 \ kJ$ $(ii)$
It follows that the heat of reaction for the following process $M + XO \rightleftharpoons MO + X$ is given by $... \ kJ$.

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The following thermochemical reactions are given:$M + \frac{1}{2} O_2 \to MO + 351.4 \ kJ$ $(i)$$X + \frac{1}{2} O_2 \to XO + 90.8 \ kJ$ $(ii)$It follows that the heat of reaction for the following process $M + XO \rightleftharpoons MO + X$ is given by $... \ kJ$.