Answer:
The enthalpy of the reaction is -2855.622 kilo Joules.
Explanation:
[tex]2C_2H_6(g) + 7O_2(g)\rightarrow 4CO_2(g) + 6H_2O(g)[/tex]
We are given:
[tex]\Delta H^o_f_{(C_{2}H_6(g))}= -84.667 kJ/mol[/tex]
[tex]\Delta H^o_f_{O_2((g))}= 0 kJ/mol[/tex]
[tex]\Delta H^o_f_{CO_2((g))}= -393.5 kJ/mol[/tex]
[tex]\Delta H^o_f_{H_2O((g))}= -241.826 kJ/mol[/tex]
The equation used to calculate enthalpy of reaction :
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=(4 mol\times \Delta H^o_f_{(CO_2(g))}+6 mol\times \Delta H^o_f_{(H_2O(g)))}-(2 mol\times \Delta G^o_f_{(C_{2}H_6(g))}+7 mol\times \Delta H^o_f_{(O_2(g)))[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[4 mol\times (-393.5 kJ/mol)+6 mol\times (-241.826 kJ/mol)]-[2 mol\times (-84.667 kJ/mol)+7 mol\times 0 kJ/mol][/tex]
[tex]=-2855.622 kJ[/tex]
The enthalpy of the reaction is -2855.622 kilo Joules.
The enthalpy change for the reaction is -2855.622 kJ/mol
Recall that enthalpy(ΔH) is a state function so;
ΔHreaction = ∑ΔHproducts - ΔHreactants
So;
The equation of the reaction is; (Recall that the question specified that we should use the smallest whole number coefficients)
2C2H6(g) + 7 O2(g) -----> 4CO2(g) + 6H2O(g)
The enthalpy of each of the reactants and products are given below;
[C2H6(g)] = −84.667 kJ/mol
O2 g = 0 KJ/mol ( O2 exists in its standard state)
[CO2(g)] = −393.5 kJ/mol
[H2O(g)] = −241.826 kJ/mol
Hence;
ΔHreaction = ∑[4 × (−393.5) + 6 × (−241.826)] - [2 × (−84.667) + (7 × 0)]
ΔHreaction = -2855.622 kJ/mol
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