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  • Calculate change in enthalpy (in kJ/mol) at 358K for the given reaction                    &

Calculate change in enthalpy (in kJ/mol) at 358K for the given reaction 

                        \small \mathrm{Fe_2O_3+3H_2\rightarrow 2Fe+3H_2O}

Given \Delta H^{\circ}_{298}=-35  KJ/mol

Substance \small \mathrm{Fe_2O_3} \small \mathrm{Fe} \small \mathrm{H_2O} \small \mathrm{H_2}
Cp (J/K.mol) 100 25 75 30

 

Option: 1

29.7


Option: 2

-29.9


Option: 3

40.6


Option: 4

-40.6


Answers (1)

best_answer

According to equation,

\mathrm{Fe_2O_3+3H_2\rightarrow 2Fe+3H_2O}

So Change in specific heat capacity ,

\mathrm{\Delta Cp =\sum Cp_{Products}-\sum Cp_{Reactants}}

\mathrm{\Delta Cp =(2\times 25 + 3 \times 75)-(100+3 \times 30)=85\: J/K.mol}

 Also, according to Kirchoff’s equation,

\mathrm{\Delta Cp =\frac{\Delta H_{T_2}-\Delta H_{T_1}}{T_2-T_1}}

\mathrm{85 \times 10^{-3} =\frac{\Delta H_{358}-(-35)}{358-298}}

\mathrm{\Delta H_{358K}=-29.9\ kJ/mol}

Posted by

Rishi

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