From the ΔE given for 25ºC, calculate ΔH in kJ at the same temperature for the...

From the ΔE given for 25ºC, calculate ΔH in kJ at the same temperature for the reaction: C2H6 (g) + 7/2 O2 (g) → 2CO2(g) + 3H2O (l) ΔE = -1553 kJ

Expert Solution

we have relation H = E + PV

dH = dE + PdV + VdP , In chemical reaction P is constant , hence dP= 0

hence dH = dE + PdV

we find PdV , In reaction total gas moles removed = product gas moles - reactanat gas moles

= 2-1-1-7/2 = - 1.5 = dn

we have PV = nRT , differneatiating both sides we get PdV + VdP = dn x R x T

here VdP = 0 , R , T are constants ,

hence PdV = dn x R x T where T = 25C = 25+273 = 298 K

= -1.5 mol x 0.08206 liter atm/molK x 298 K

= -36.7 atmliter

= -36.7 x 101.325 J ( 1atmliter = 101.325 J)

= -3717 J = -3.717 KJ

now dH = dE + PdV

dH = -1553 KJ - 3.717 KJ

= - 1556.7 KJ

queen_honey_blossom answered 2 years ago

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