What is the warm of burning of ethane, C2H6, in kilojoules every mole of ethane?

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Answer

The warmth of reaction left(Delta H_mathrmr m ight) is the difference between sum of heat of development of products and also sum of warmth of formation of reactants.

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The warm of reaction for combustion left(Delta H_mathrmrm ight)_ ext combution is nothing yet heat of combustion.

The well balanced combustion reaction equation that ethane is,


mathrmC_2 mathrmH_6(g)+frac72 mathrmO_2(g) ightarrow 2 mathrmCO_2(g)+3 mathrmH_2 mathrmO(l)

From the data, the enthalpy of development at 25^circ mathrmC, that is, traditional enthalpy of formation left(Delta H_f^circ ight), of every component is

left(Delta H_f ight)_mathrmc_2 mathrmH_mathrmz(g)=-83.820 mathrm~kJ / mathrmmol,left(Delta H_f ^circ ight)_mathrmO_2(g)=0 mathrm~kJ / mathrmmol

left(Delta H_f^circ ight)_mathrmco_2(g)=-393.509 mathrm~kJ / mathrmmol, and left(Delta H_f^circ ight)_mathrmH_2 mathrmO(b)=-285.830 mathrm~kJ / mathrmmol

Since, the heat of reaction left(H_mathrmrm ight) is,


Delta H_mathrmrm=left(sum_i m_ileft(Delta H_f ight)_i ight)_ ext product -left(sum_j n_jleft(Delta H_f ight)_j ight)_mathrmreactant

Here, subscripts i and j suggests component, m_i is the moles of the product i, and also n_j is the mole of reactant j.

Re-write the over equation for the existing reaction as,

eginequationleft(Delta H_mathrmrm ight)_ ext combution =left(m_mathrmCO_2(g)left(Delta H_f ^circ ight)_mathrmCO_2(g)+m_mathrmH_2 mathrmO(t)left(Delta H_f ^0 ight)_mathrmH_2 0(b) ight)-left(eginarrayln_mathrmC_2 mathrmH_6(g)left(Delta H_f ^circ ight)_mathrmC, mathrmH_6(g) \+n_mathrmO_2(mathrm~g)left(Delta H_f circ ight)_0_2(g)endarray ight)endequationObserve the reaction, which is clear that m_mathrmco_2(g)=2 mathrm~mol ; m_mathrmH_2 mathrmO(mathrmg)=3 mathrm~mol ; n_mathrmc_2 mathrmH_2(g)=1 mathrm~mol and also n_0_2(g)=frac72 mathrm~mol

Substitute the equivalent numerical values to the variables in the over equation.

eginequationeginalignedleft(Delta H_mathrmrm ight)_ ext combition &=left(eginarrayl2 mathrm~mol imes-393.509 mathrm~kJ / mathrmmol+ \3 mathrm~m circ mathrml imes-285.83 mathrm~kJ / mathrmm circ 1endarray ight)-left(eginarraycoperatornamelm circ 1 imes-83.820 mathrm~kJ / mathrmmol \+frac72 mathrm~mol imes 0 mathrm~kJ / mathrmmolendarray ight) \=&-1560.688 mathrm~kJ / mathrmmol quadleft\beginarrayl1- ext 'sign suggests that heat is liberated as result of the \ ext reaction endarray ight\endalignedendequationHere, the worth of left(Delta H_mathrmrm ight)_ ext ccmbutim is calcul ated for burning of 1 mathrm~mol the ethane.

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Symbolically, left(Delta H_ ext ran ight)_ ext combution is -1560.688 mathrm~kJ / mathrmmol the mathrmCH_3 mathrmOH combusted.