Use the molar bond enthalpy data in the table to estimate the value of ΔH°rxn for the equation:?
The answer is -608 kJ/mol. The negative is significant because it shows that the reaction is exothermic. You were both correct to total the bond strengths before and after the reaction. However, remember that what is shown on the table is NOT stored energy; it is the energy required to break the bond. (Think of this as a "negative" potential energy, similar to the energy required to rescue someone from a hole) Before: Reactant bond strength = 1634 kJ/mol This means that it requires 1634 kJ/mol to break the bonds. After: Product bond strength = 2242 kJ/mol It requires a greater amount of energy to break the bonds of the products. This means that energy was released during the reaction.
Use bond energy values to estimate delta H?
The bond energy is the energy required to break up a bound. The energy to from a bond is the negative bond energy. Hence the heat of reaction is approximately the sum of the bond energies of the reactants minus the sum of the bond energies of the products. a) H2 + Cl2 → 2 HCl The bonds are H2 one H-H bond Cl2 one Cl-Cl bond HCl one H-Cl bond Hence: ΔH = D(H-H) + D(Cl-Cl) - 2·D(H-Cl) = 432kJ/mol + 240kJ/mol - 2 · 428kJ/mol = -184kJ/mol b) N2 + 3 H2 → 2 NH3 The bonds are N2 one N≡N bond (triple bond) H2 one H-H bond NH3 three N-H bonds Hence: ΔH = D(N≡N) + 3·D(H-H) - 2·3·D(N-H) = 942kJ/mol + 3·432kJ/mol - 6·386kJ/mol = -78kJ/mol
Calculate the avg. N-H bond enthalpy in NH3:?
To find the bond enthalpy, you subtract the energy of the bonds formed (since they release energy) and add the energy of the bonds broken (since you need to supply energy). 1/2 an N2 bond is 941/2 = 470.5 kJ/mol 3/2 H2 bonds = 3×436/2 = 654 kJ/mol Total energy = bonds broken-bonds formed = (470.5 + 654) - 3(N-H) Total energy = -46.3 kJ/mol Therefore, -3(N-H) + 1124.5 = -46.3 3(N-H) = 1124.5 + 46.3 = 1170.8 (N-H) = 1170.8/3 = 390.2667 ≈ 390 kJ/mol Remember: total energy of reaction = (total energy of bonds broken) - (total energy of bonds formed) Hope that helps you!
How do you calculate the standard enthalpy of formation of acetylene from the heat of combustion of C2H2,C (graphite) and H2 given as -1300kj/mol, 395kj/mol and -286kj/mol respectively?
You usually calculate the enthalpy change of combustion from enthalpies of formation.The standard enthalpy of combustion is [math]ΔH_c^∘[/math].It is the heat evolved when 1 mol of a substance burns completely in oxygen at standard conditions. For example,[math]C_{2}H_{2}(g)+\dfrac{5}{2}O_{2}(g)→2CO_{2}(g)+H_{2}O(l)[/math]You calculate [math]ΔH_c^∘ [/math]from standard enthalpies of formation:[math]ΔH_c^∘=∑ΔH_f^∘(p)−∑ΔH_f^∘(r)[/math]where, [math]p[/math]stands for "products" and [math]r [/math]stands for "reactants".For each product, you multiply its [math]ΔH_f^∘ [/math]by its coefficient in the balanced equation and add them together.Do the same for the reactants. Subtract the reactant sum from the product sum.EXAMPLE:Use the following enthalpies of formation to calculate the standard enthalpy of combustion of acetylene, [math]C_{2}H_{2}[/math].[math]C_{2}H_{2}(g)+\dfrac{5}{2}O_{2}(g)→2CO_{2}(g)+H_{2}O(l)[/math][math]C_{2}H_{2}(g)=226.73 kJ/mol ; [/math][math]{ΔH_{CO_2}^o}(g)=-393.5 kJ/mol ;[/math][math]{ΔH_{H_2{O}}^o}(l)=-285.8 kJ/mol[/math]Solution:[math]C_{2}H_{2}(g)+\dfrac{5}{2}O_{2}(g)→2CO_{2}(g)+H_{2}O(l)[/math][math]ΔH_c^∘=∑ΔH_f^∘(p)−∑ΔH_f^∘(r)[/math][math][2 × (-393.5) + (-295.8)] – [226.7 + 0] kJ=-1082.8 - 226.7= -1309.5 kJ[/math]The heat of combustion of acetylene is [math]-1309.5 kJ/mol.[/math]
Bond enthalpy help?
Ammonia is produced directly from nitrogen and hydrogen by using the Haber process. The chemical reaction is: N2(g)+3H2(g)→2NH3(g) ---------------------------------------... Q: Use bond enthalpies to estimate the enthalpy change for the reaction. Express your answer using two significant figures? Q: Calculate the true enthalpy change as obtained using ΔH∘f values. PS: The answers are in units of: kJ/2molNH3 -Can some one explain and give me the steps throughout please?
Calculate the enthalpy for the combustion of pentane?
The combustion of pentane, C5H12, occurs via the reaction: C5H12 (g) + 8O2 (g) ---> 5CO2 (g) + 6H2O (g) with heat of formation values given by the following table: Substance: Delta H (f) (kJ/mol): C5H12 (g) -35.1 CO2 (g) -393.5 H2O (g) -241.8 Calculate the enthalpy for the combustion of pentane