calculate the standard reaction enthalpy for the reaction between clf3 and o2. 2clf3(g) + 2o2(g) → cl2o(g) +…

calculate the standard reaction enthalpy for the reaction between clf3 and o2. 2clf3(g) + 2o2(g) → cl2o(g) + 3of2(g) compound δh°f (kj/mol) clf3 -163.2 cl2o 80.3 of2 24.5 δhrxn = ? kj enter either a + or - sign and the magnitude. δh,rxn (kj)

calculate the standard reaction enthalpy for the reaction between clf3 and o2. 2clf3(g) + 2o2(g) → cl2o(g) + 3of2(g) compound δh°f (kj/mol) clf3 -163.2 cl2o 80.3 of2 24.5 δhrxn = ? kj enter either a + or - sign and the magnitude. δh,rxn (kj)

Answer

Answer:

+538.4

Explanation:

Step1: Recall the formula

$\Delta H_{rxn}=\sum n\Delta H^{\circ}{f}(products)-\sum m\Delta H^{\circ}{f}(reactants)$

Step2: Calculate products' $\Delta H^{\circ}_{f}$ sum

$n_1\Delta H^{\circ}{f}(Cl_2O)+n_2\Delta H^{\circ}{f}(OF_2)=1\times80.3 + 3\times24.5=80.3+73.5 = 153.8$

Step3: Calculate reactants' $\Delta H^{\circ}_{f}$ sum

$m_1\Delta H^{\circ}{f}(ClF_3)+m_2\Delta H^{\circ}{f}(O_2)=2\times(- 163.2)+2\times0=-326.4$

Step4: Calculate $\Delta H_{rxn}$

$\Delta H_{rxn}=153.8-(-326.4)=153.8 + 326.4=538.4$