Consider the structure of 1, 4-pentadiene. Write out a balanced chemical reactio
ID: 517332 • Letter: C
Question
Consider the structure of 1, 4-pentadiene. Write out a balanced chemical reaction for the heat of formation of 1, 4 pentadiene (delta H_c degree = -769 kcal mol^-1) Given Delta H_f degree ((E)-1, 3 - pentadiene) = 18 kcal mol^-1, determine the Delta H_c * of (E) - 1, 3 - pentadiene. -1, 3 - pentadiene is more stable than 1, 4 pendtadiene, an isomer. Provide a plausible explanation. Estimate Delta _C H degree of (Z) - 1, 3 pentadiene relative to (E) - 1, 3 pentadiene? Briefly explain. Given the Delta H_f degree values indicated, rank the compounds below from most heat released upon combustion (most negative Delta H_c degree) to least heat released. Briefly explain you reasoning.Explanation / Answer
(1) The formation reaction for 1,4 pentadiene from its constituent elements is given as
5C(s) + 4H2(g) ------> C5H8 (l)
The enthalpy of the reaction is given as:
Hrxn = [Hf(C5H8 (l))] - [4Hf(H2(g)) + 5Hf(C(s))]
From the thermodynamics property table we know that
Hf(C(s)) = 0 kJ mol-1; Hf(H2(g)) = 218 kJ mol-1;
Hf(C5H8 (l)) = 25 kcal mol-1 …. Given
1 kcal = 4.184 kJ
Hf(C5H8 (l)) = 25 X 4.184 kJ mol-1 = 104.6 kJ mol-1
Hrxn = [104.6] - [4(218) + 5(0)]
= 104.6 - 872
Hrxn = -767.4 kJ (exothermic)
(2) The combustion reaction for 1,4 pentadiene is given as
C5H8 (l) + 7O2(g) -------> 5CO2(s) + 4H2O(g)
The enthalpy of the reaction is given as:
Hrxn = [4Hf(H2O(g)) + 5Hf(CO2(s))]- [Hf(C5H8 (l)) + 7Hf(O2(s))]
From the thermodynamics property table we know that
Hf(O2(g)) = 0 kJ mol-1; Hf(H2O(g)) = -241.8 kJ mol-1; Hf(CO2(g)) = -393.5 kJ mol-1
Hf(C5H8 (l)) = -769 kcal mol-1 …. Given
1 kcal = 4.184 kJ
Hf(C5H8 (l)) = -769 X 4.184 kJ mol-1 = -3217.5 kJ mol-1
Hrxn = [4(-241.8) + 5(-393.5)] - [-3217.5 + 7(0)]
= -967.2 – 1967.5 + 3217.5
Hrxn = 282.8 kJ
(3)
The formation reaction for 1,3 pentadiene from its constituent elements is given as
5C(s) + 4H2(g) ------> C5H8 (l)
Hf(C5H8 (l)) = 18 kcal mol-1 …. Given
1 kcal = 4.184 kJ
Hf(C5H8 (l)) = 18 X 4.184 kJ mol-1 = 75.3 kJ mol-1
By inverting the reaction
C5H8 (l) ------> 5C(s) + 4H2(g)
Hf(C5H8 (l)) = -75.3 kJ mol-1
The formation of CO2 is given as
5C + 5O2 --> 5CO2
The formation of H2O is given as
4H2 + 2O2 --> 4H2O
Adding the above three equations we get the combustion reaction for 1,3 pentadiene and it is given as
C5H8 (l) + 7O2(g) -------> 5CO2(s) + 4H2O(g)
Hf(H2O(g)) = -241.8 kJ mol-1; Hf(CO2(g)) = -393.5 kJ mol-1
Hc = Hf(C5H8 (l)) + 4Hf(H2O(g)) + 5Hf(CO2(g))
= -75.3 + 4(-241.8) + 5(-393.5)
=-75.3 -967.2 – 1967.5 = 3010 kJ
(4) The formation reaction for 1,3 pentadiene from its constituent elements is given as
5C(s) + 4H2(g) ------> C5H8 (l)
The enthalpy of the reaction is given as:
Hrxn = [Hf(C5H8 (l))] - [4Hf(H2(g)) + 5Hf(C(s))]
From the thermodynamics property table we know that
Hf(C(s)) = 0 kJ mol-1; Hf(H2(g)) = 218 kJ mol-1;
Hf(C5H8 (l)) = 18 kcal mol-1 …. Given
1 kcal = 4.184 kJ
Hf(C5H8 (l)) = 18 X 4.184 kJ mol-1 = 75.3 kJ mol-1
Hrxn = [75.3] - [4(218) + 5(0)]
= 75.3 - 872
Hrxn = -796.7 kJ (exothermic)
1,3 pentadiene is more stable than 1,4 pentadiene because
(i)The Hrxn is lower for 1,3 pentadiene (-796.7 kJ) when compared to 1,4 pentadiene (-767.4 kJ).
(ii)1,3 pentadiene is a conjugated system (alternate double bonds) and the resonance adds stability to the compound.
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