Estimate the standard enthalpy change (Delta H_(can)^circ ) of the reaction D using reactions A-C A) NH_(3)arrow N_(2)+H_(2)Delta H_(rxn)^circ =+92kJ/mol B) H_(2)Oarrow H_(2)+O_(2)Delta H_(rxn)^circ =+572kJ/mol C) N_(2)+O_(2)arrow NO Delta H_(oun)^circ =-180kJ/mol D) NH_(3)+O_(2)arrow NO+H_(2)O Delta H_(rm)^circ -? 5) How much heat can be produced from a reaction mixture of 50 g of iron (III) oxide and aluminium in the thermite reaction:

To estimate the standard enthalpy change $(\Delta H_{can}^{\circ })$ of the reaction D using reactions A-C, we can use Hess's Law. According to Hess's Law, the total enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps that lead to the overall reaction.<br /><br />Given reactions:<br />A) $NH_{3}\rightarrow N_{2}+H_{2}\quad \Delta H_{rxn}^{\circ }=+92kJ/mol$<br />B) $H_{2}O\rightarrow H_{2}+O_{2}\quad \Delta H_{rxn}^{\circ }=+572kJ/mol$<br />C) $N_{2}+O_{2}\rightarrow NO\quad \Delta H_{rxn}^{\circ }=-180kJ/mol$<br /><br />We want to find the enthalpy change for reaction D: $NH_{3}+O_{2}\rightarrow NO+H_{2}O$<br /><br />To do this, we need to manipulate the given reactions to match the desired reaction. We can reverse reaction A and multiply it by 2, reverse reaction B and multiply it by 2, and keep reaction C as it is.<br /><br />Reversed and multiplied reactions:<br />A') $N_{2}+H_{2}\rightarrow NH_{3}\quad \Delta H_{rxn}^{\circ }=-92kJ/mol$<br />B') $H_{2}+O_{2}\rightarrow H_{2}O\quad \Delta H_{rxn}^{\circ }=-572kJ/mol$<br />C) $N_{2}+O_{2}\rightarrow NO\quad \Delta H_{rxn}^{\circ }=-180kJ/mol$<br /><br />Now, we can add these manipulated reactions together to obtain the desired reaction D:<br /><br />A') $N_{2}+H_{2}\rightarrow NH_{3}\quad \Delta H_{rxn}^{\circ }=-92kJ/mol$<br />B') $H_{2}+O_{2}\rightarrow H_{2}O\quad \Delta H_{rxn}^{\circ }=-572kJ/mol$<br />C) $N_{2}+O_{2}\rightarrow NO\quad \Delta H_{rxn}^{\circ }=-180kJ/mol$<br /><br />Adding these reactions together, we get:<br />$N_{2}+H_{2}+H_{2}+O_{2}+N_{2}+O_{2}\rightarrow NH_{3}+H_{2}O+NO$<br /><br />Simplifying, we have:<br />$2N_{2}+3H_{2}+O_{2}\rightarrow 2NH_{3}+H_{2}O+NO$<br /><br />This is the desired reaction D. Now, we can calculate the enthalpy change for reaction D by adding the enthalpy changes of the manipulated reactions:<br /><br />$\Delta H_{can}^{\circ } = \Delta H_{A'}^{\circ } + \Delta H_{B'}^{\circ } + \Delta H_{C}^{\circ }$<br />$\Delta H_{can}^{\circ } = (-92kJ/mol) + (-572kJ/mol) + (-180kJ/mol)$<br />$\Delta H_{can}^{\circ } = -844kJ/mol$<br /><br />Therefore, the standard enthalpy change $(\Delta H_{can}^{\circ })$ of the reaction D is -844 kJ/mol.<br /><br />Regarding the thermite reaction, it is a highly exothermic reaction between iron (III) oxide and aluminum. The balanced chemical equation for the thermite reaction is:<br /><br />$Fe_{2}O_{3}+2Al\rightarrow 2Fe+Al_{2}O_{3}$<br /><br />To calculate the amount of heat produced from the reaction mixture, we need to know the amount of iron (III) oxide and aluminum present in the mixture. However, the given information only provides the masses of iron (III) oxide and aluminum, which is not enough to calculate the heat produced. We would need additional information such as the specific heat capacities of the substances involved or the reaction conditions to determine the heat produced.