Question

Based on enthalpy of formation data species ∆H◦ f H2S(g) −20.63 kJ/mol O2(g) 0 kJ/mol H2O(ℓ) −285.83 kJ/mol SO2(g) −296.83 kJ/mol calculate ∆Hrxn for 2 H2O(ℓ) + 2 SO2(g) ←→ 2 H2S(g) + 3 O2(g) 1. 562 kJ · mol−1 2. −562 kJ · mol−1 3. −1124 kJ · mol−1 4. 1124 kJ · mol−1

Answer 1

Answer: The enthalpy change of the reaction is 1124 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{\text{(product)}}]-\sum [n\times \Delta H^o_f_{\text{(reactant)}}]$

For the given chemical reaction:

$2H_2O(l)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(H_2S(g))})+(3\times \Delta H^o_f_{(O_2(g))})]-[(2\times \Delta H^o_f_{(H_2O(l))})+(2\times \Delta H^o_f_{(SO_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.83kJ/mol\\\Delta H^o_f_{(H_2S(g))}=-20.63kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-20.63))+(3\times (0))]-[(2\times (-285.83))+(2\times (-296.83))]\\\\\Delta H_{rxn}=1124kJ$

Hence, the enthalpy change of the reaction is 1124 kJ