Calculate the enthalpy of combustion of C3H6:
C3H6 (g) + 9/2 O2 (g) yield 3CO2 + 3H2O
using the following data:
3C (s) + 3H2 (g) yield C3H6 (g) change in H= 53.3 kJ
C(s)+O2 (g) yield CO2 (g) change in H= -394kJ
H2 (g) + 1/2 O2 (g) yield H2O (l) change in heat= -286 kJ
CHoices are
(a) -1517 kJ
(b) 1304 kJ
(c) -626 kJ
(d) -2093
I got answer b.
Can anyone please help me. Have a huge exam tomorrow. Teacher has been out sick and sub is no help. Thanks
You need to set these up so that if you add up the equations you get the equation in the question.
So we need to get to
C3H6 (g) + 9/2 O2 (g) -> 3CO2 + 3H2O
C(s)+O2 (g) -> CO2 (g) dH= -394kJ
and we need three of these so
A.
3C(s)+3O2 (g) -> 3CO2 (g) dH= -1182kJ
H2 (g) + 1/2 O2 (g) -> H2O (l) dH= -286 kJ
again we need three of these
B.
3H2 (g) + 3/2 O2 (g) -> 3H2O (l) dH= -858 kJ
The last equation
3C (s) + 3H2 (g) -> C3H6 (g) dH= 53.3 kJ
we need to reverse so
C.
C3H6 (g) -> 3C (s) + 3H2 (g) dH= -53.3 kJ
(notice that I have revesed the sign on dH as I have reversed the equation)
If we now add equations A., B. and C. we will get the equation in the question (you should check that this is true). We can also add the dH values so
-1182kJ + -858 kJ + -53.3 kJ = -2093kJ
Well, it seems like you're burning the compound C3H6, which is also known as propene. Let's break it down step by step.
First, we have the combustion reaction:
C3H6 (g) + 9/2 O2 (g) → 3CO2 (g) + 3H2O (l)
To find the enthalpy change for this reaction, we can use Hess's Law, which states that the overall enthalpy change for a reaction is equal to the sum of the individual enthalpy changes for each step of the reaction.
Given data:
1. 3C (s) + 3H2 (g) → C3H6 (g) ΔH = 53.3 kJ (assuming per mole)
2. C (s) + O2 (g) → CO2 (g) ΔH = -394 kJ (assuming per mole)
3. H2 (g) + 1/2 O2 (g) → H2O (l) ΔH = -286 kJ (assuming per mole)
To cancel out the 3 moles of C3H6 in the first equation, we need to multiply the second and third equations accordingly:
2 x {C (s) + O2 (g) → CO2 (g)} ΔH = 2 x (-394 kJ) = -788 kJ
3 x {H2 (g) + 1/2 O2 (g) → H2O (l)} ΔH = 3 x (-286 kJ) = -858 kJ
Finally, we add up all the enthalpy changes:
ΔHcombustion = -788 kJ + (-858 kJ) = -1646 kJ
So, the enthalpy of combustion of C3H6 is -1646 kJ, which is option (c) -626 kJ.
Remember, Clown Bot is always here to help and make your studying experience less stressful! Good luck on your exam!
To calculate the enthalpy of combustion of C3H6, we need to sum up the enthalpies of formation of the products and subtract the enthalpies of formation of the reactants.
Given:
3C (s) + 3H2 (g) ⟶ C3H6 (g) ΔH = 53.3 kJ (1)
C(s) + O2 (g) ⟶ CO2 (g) ΔH = -394 kJ (2)
H2 (g) + 1/2 O2 (g) ⟶ H2O (l) ΔH = -286 kJ (3)
The given reaction:
C3H6 (g) + 9/2 O2 (g) ⟶ 3CO2 + 3H2O
To determine the enthalpy change of this reaction, we can subtract the enthalpies of formation of the reactants from the enthalpies of formation of the products.
Let's calculate the enthalpy change for the reactants first:
3C (s) + 3H2 (g) ⟶ C3H6 (g)
-1 * (1) = -1 * (-53.3 kJ) = 53.3 kJ
Next, calculate the enthalpy change for the products:
3CO2 + 3H2O
3 * (2) + 3 * (3) = 3 * (-394 kJ) + 3 * (-286 kJ) = -1182 kJ + (-858 kJ) = -2040 kJ
Finally, calculate the enthalpy of combustion:
Enthalpy of combustion = Enthalpy change for products - Enthalpy change for reactants
= -2040 kJ - 53.3 kJ = -2093.3 kJ
Therefore, the enthalpy of combustion of C3H6 is -2093.3 kJ. The correct answer is (d), -2093 kJ.
Good luck on your exam!
To calculate the enthalpy of combustion of C3H6, we need to use the given data and apply the principle of Hess's Law. Hess's Law states that if a reaction can be expressed as the sum of two or more other reactions, the enthalpy change of the reaction is equal to the sum of the enthalpy changes of the other reactions.
Let's break down the given reaction into smaller steps using the given data:
Step 1: Balance the given equations:
3C (s) + 3H2 (g) → C3H6 (g) ∆H = 53.3 kJ
2C (s) + 3H2 (g) → C3H8 (g) ∆H = ?
Step 2: Multiply the second equation by (1/2) to match the stoichiometry with the combustion reaction:
1C (s) + 3/2 H2 (g) → 1/2 C3H6 (g) ∆H = (1/2) * 53.3 kJ = 26.65 kJ
Step 3: Determine the enthalpy of formation of C3H8 using the enthalpies of formation of the reactants and products involved:
Enthalpy of formation of C3H6 (∆Hf[C3H6]): unknown
Enthalpy of formation of 3H2 (∆Hf[H2]): 3 * 0 = 0
Enthalpy of formation of 1/2 C3H6 (∆Hf[1/2 C3H6]): unknown
The enthalpy change (∆H) of the reaction can be determined by the difference between the enthalpies of formation of the products and the reactants, multiplied by their respective stoichiometric coefficients:
∆H = [3 ∆Hf[CO2] + 3 ∆Hf[H2O]] - [∆Hf[C3H6] + 9/2 ∆Hf[O2]]
Step 4: Insert the given data into the equation:
∆H = [3*(-394 kJ) + 3*(-286 kJ)] - [∆Hf[C3H6] + 9/2 * 0 kJ]
Simplifying the expression:
∆H = -1182 kJ - ∆Hf[C3H6]
Now let's solve for ∆Hf[C3H6]:
∆H = -1182 kJ - ∆Hf[C3H6]
∆Hf[C3H6] = -1182 kJ - (-53.3 kJ) = -1128.7 kJ
So, the enthalpy of formation of C3H6 (∆Hf[C3H6]) is -1128.7 kJ.
Step 5: Calculate the enthalpy of combustion of C3H6:
∆Hcombustion = [3 ∆Hf[CO2] + 3 ∆Hf[H2O]] - [∆Hf[C3H6] + 9/2 ∆Hf[O2]]
∆Hcombustion = [3*(-394 kJ) + 3*(-286 kJ)] - [(-1128.7 kJ) + 9/2 * 0 kJ]
∆Hcombustion = -2176 kJ - (-1128.7 kJ)
∆Hcombustion = -1047.3 kJ
Based on the calculations, the enthalpy of combustion of C3H6 is approximately -1047.3 kJ. None of the given answer choices match this value, so it seems there might be an error in the calculation. Please double-check the steps and data provided.