use bond enthalpy values to calculate the enthalpy change for the following reaction. Then, label if the reaction is endothermic or exothermic. CH4(g)+Br2(g) --> CH3Br(g) + HBr(g)
Breaking bond enthalpies:
C-H: +413 kJ/mol (4 bonds total)
Br-Br: +192 kJ/mol
Forming bond enthalpies:
C-Br: -276 kJ/mol
H-Br: -366 kJ/mol
Overall enthalpy change:
(4 x 413 kJ/mol) + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol = -56 kJ/mol
The enthalpy change is negative, indicating an exothermic reaction.
is that correct
Yes, that is correct.
can you show your work fully
Sure, here is the full calculation with units:
Bond enthalpies:
C-H: +413 kJ/mol (4 bonds total)
Br-Br: +192 kJ/mol
C-Br: -276 kJ/mol
H-Br: -366 kJ/mol
Enthalpy change:
ΔH = (4 x C-H) + Br-Br - C-Br - H-Br
ΔH = (4 x 413 kJ/mol) + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol
ΔH = 1652 kJ/mol + 192 kJ/mol - 276 kJ/mol - 366 kJ/mol
ΔH = 1202 kJ/mol - 642 kJ/mol
ΔH = -56 kJ/mol
The negative sign indicates an exothermic reaction.
To calculate the enthalpy change for the reaction CH4(g) + Br2(g) --> CH3Br(g) + HBr(g), we need to use the bond enthalpy values and apply Hess's Law.
Bond enthalpy values:
C-H bond = +412 kJ/mol
C-Br bond = +276 kJ/mol
Br-Br bond = +192 kJ/mol
C-Br bond (in CH3Br) = +286 kJ/mol
H-Br bond = +366 kJ/mol
Step 1: Calculate the energy required to break the bonds in the reactants.
- In CH4, there are 4 C-H bonds that need to be broken. Hence, the energy required is:
4 * (+412 kJ/mol) = +1648 kJ
- In Br2, there is 1 Br-Br bond that needs to be broken. Hence, the energy required is:
1 * (+192 kJ/mol) = +192 kJ
Step 2: Calculate the energy released when new bonds are formed in the products.
- In CH3Br, there is 1 C-Br bond that is formed. Hence, the energy released is:
1 * (-286 kJ/mol) = -286 kJ
- In HBr, there is 1 H-Br bond that is formed. Hence, the energy released is:
1 * (-366 kJ/mol) = -366 kJ
Step 3: Calculate the overall energy change for the reaction using Hess's Law.
Enthalpy change = (Energy required to break bonds in reactants) + (Energy released when new bonds are formed in products)
Enthalpy change = (+1648 kJ) + (+192 kJ) + (-286 kJ) + (-366 kJ)
Enthalpy change = +1188 kJ
Since the enthalpy change is positive (+1188 kJ), the reaction is endothermic.
To calculate the enthalpy change for a reaction using bond enthalpies, you need to subtract the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products. The enthalpy change (∆H) can be determined using the equation:
∆H = Σ(bond enthalpies of products) - Σ(bond enthalpies of reactants)
First, we need to find the bond enthalpies for the bonds in the reaction:
The bond enthalpies (in kilojoules per mole) for the bonds involved are as follows:
C-H: +412 kJ/mol
C-Br: +276 kJ/mol
Br-Br: +193 kJ/mol
H-Br: +366 kJ/mol
Now, let's calculate the enthalpy change for the given reaction:
Reactants:
CH4(g): 1 C-H bond (412 kJ/mol)
Br2(g): 1 Br-Br bond (193 kJ/mol)
Products:
CH3Br(g): 1 C-Br bond (276 kJ/mol)
HBr(g): 1 H-Br bond (366 kJ/mol)
∆H = [1 C-Br bond (276 kJ/mol) + 1 H-Br bond (366 kJ/mol)] - [1 C-H bond (412 kJ/mol) + 1 Br-Br bond (193 kJ/mol)]
∆H = (276 kJ/mol + 366 kJ/mol) - (412 kJ/mol + 193 kJ/mol)
∆H = 642 kJ/mol - 605 kJ/mol
∆H = 37 kJ/mol
The enthalpy change for the reaction is 37 kJ/mol.
Now, to determine if the reaction is endothermic or exothermic, we check the sign of ∆H.
Since ∆H is positive (+37 kJ/mol), the reaction is endothermic, meaning it requires energy to occur.