Calculate the change in enthalpy of a reaction, Delta H_rxn, using the enthalpie

Question

Calculate the change in enthalpy of a reaction, Delta H_rxn, using the enthalpies of formation, Delta H_f, for the reactants and products. Use the equation: Delta H_rxn = 2(Delta H_f-products) - sigma (Delta H_f-reactants) a. Use the standard heats of formation provided to determine Delta H_rxn for the following reaction: CH_4 (g) + 4 Cl_2 (g) rightarrow CCl_4 (g) + 4 HCI (g) Delta H_rxn = ? b. Use the standard heats of formation provided to determine Delta H_rxn for the following reaction: SO_2CI_2 (I) + 2 H_2O (I) rightarrow 2 HCI (g) + H_2SO_4 (I) Delta H_rxn = ? c. Use the standard heats of formation provided to determine Delta H_rxn for the following reaction: 3 Fe_2O_3 (s) + CO (g) rightarrow 2 Fe_3O_4 (s) + CO_2 (g) Delta H_rxn = ?

Explanation / Answer

a.delta Hrxn = sum of delta Hf for products - sum of delta Hf for reactants

= [deltaHf(CCl4) + 4*deltaHf(HCl)]-[deltaHf(CH4)+ 4*deltaHf(Cl2)]

= [-96 +(4*-92.3)] - [-75 +4*0]

= -465.2 + 75 = -390.2kJ/mol

b. delta Hrxn = [ 2*deltaHf(HCl) + deltaHf(H2SO4)] - [deltaHf(SO2Cl2) + 2*deltaHf(H2O)]

= [2*(-92.3) + (-81.4)] - [-254.6 + 2*(-285.8)]

= -266-(-826.2) = 560.2kJ/mol

c.DeltaHrxn = [2*deltaHf(Fe3O4) + Hf(CO2)]-[3*deltaHf(Fe2O3) + deltaHf(CO)]

= [ 2*(-1118) + (-394)] - [ 3*(-824) + (-111)

= -2630 - (-2583) = -42kJ/mol

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