Question

Use bond energies to calculate δhrxn for the reaction. n2(g)+3cl2(g)→2ncl3(g)

Answer 1

ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)

Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹

According to the balanced equation,
ΣδΗ(bond breaking)  = N ≡ N x 1 + Cl - Cl x 3
                                  = 945 + 3(242)
                                  = 1671 kJ mol⁻¹
ΣδΗ(bond making)    = N - Cl x 3 x 2
                                 = 192 x 6
                                 = 1152 kJ mol⁻¹

δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
          = 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
          = 519 kJ mol⁻¹

Answer 2

The value of $\Delta {H_{{\text{reaction}}}}$ of the reaction ${{\text{N}}_{\text{2}}}\left( g \right) + 3{\text{C}}{{\text{l}}_2}\left( g \right) \to 2{\text{NC}}{{\text{l}}_3}\left( g \right)$ is $\boxed{519\;{\text{kJ/mol}}}$.

Further explanation:

Heat of reaction:

The heat released or absorbed in a chemical reaction due to the difference in the bond energies (BE) of reactants and products in the reaction is known as the heat of reaction. It is represented by$\Delta {H_{{\text{reaction}}}}$.

The heat of reaction $\left( {\Delta {H_{{\text{reaction}}}}} \right)$can have two values:

Case I: If the reaction is endothermic, more energy needs to be supplied to the system than that released by it. So $\Delta {H_{{\text{reaction}}}}$ comes out to be positive.

Case II: If the reaction is exothermic, more energy is released by the system than that supplied to it. So $\Delta {H_{{\text{reaction}}}}$ comes out to be negative.

The formula to calculate the heat of reaction is,

$\boxed{\Delta {H_{{\text{reaction}}}} = \sum {\text{B}}{{\text{E}}_{{\text{reactant bond broken}}}} - \sum {\text{B}}{{\text{E}}_{{\text{product bond formed}}}}}$

Here,

$\Delta {H_{{\text{reaction}}}}$ is the heat of reaction.

${\text{B}}{{\text{E}}_{{\text{product bond formed}}}}$ is the bond energy of bond formation in products.

${\text{B}}{{\text{E}}_{{\text{reactant bond broken}}}}$ is the bond energy of bond breakage in reactants.

The given reaction occurs as follows:

${{\text{N}}_{\text{2}}}\left( g \right) + 3{\text{C}}{{\text{l}}_2}\left( g \right) \to 2{\text{NC}}{{\text{l}}_3}\left( g \right)$

The number of broken bonds is 1 N-N triple bond and 3 Cl-Cl bonds.

The number of bonds formed is 6 N-Cl bonds.

The formula to calculate the enthalpy of the given reaction is as follows:

$\Delta {H_{{\text{reaction}}}} = \left[ {\left( {1{\text{B}}{{\text{E}}_{{\text{N}} - {\text{N}}}} + 3{\text{B}}{{\text{E}}_{{\text{Cl}} - {\text{Cl}}}}} \right) - \left( {{\text{6B}}{{\text{E}}_{{\text{N}} - {\text{Cl}}}}} \right)} \right]$                   …… (1)

The bond energy of N-N triple bond is 945 kJ/mol.

The bond energy of N-Cl bond is 192 kJ/mol.

The bond energy of Cl-Cl bond is 242 kJ/mol.

Substitute 945 kJ/mol for${\text{B}}{{\text{E}}_{{\text{N}} - {\text{N}}}}$, 192 kJ/mol for${\text{B}}{{\text{E}}_{{\text{N}} - {\text{Cl}}}}$, 242 kJ/mol for${\text{B}}{{\text{E}}_{{\text{Cl}} - {\text{Cl}}}}$, in equation (1).

$\Delta {H_{{\text{reaction}}}}&=\left[ {\left( {\left( {{\text{1 mol}}} \right)\left( {945\;{\text{kJ/mol}}} \right) + \left( {{\text{3 mol}}} \right)\left( {242\;{\text{kJ/mol}}} \right)} \right) - \left( {{\text{6 mol}}} \right)\left( {192\;{\text{kJ/mol}}} \right)} \right]\\\begin{aligned}&=\left[{\left({94{\text{5 kJ/mol}} + 72{\text{6 kJ/mol}}} \right) - 115{\text{2 kJ/mol}}} \right]\\&=\left[{167{\text{1 kJ/mol}} - 115{\text{2 kJ/mol}}} \right] \\ \end{aligned}$

Learn more:

1. Calculate the enthalpy change using Hess’s Law: brainly.com/question/11293201

2. Find the enthalpy of decomposition of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: N2, Cl2, NCl3, bond energies, N-Cl bond, N-N bond, Cl-Cl bond, 519 kJ/mol, heat of reaction, released, absorbed, exothermic, endothermic, positive, negative.