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Hydrogen (H2) combines with chlorine (Cl2) to form hydrochloric acid (HCl): H2 + Cl2 → 2HCl. Using the chart below, what is the estimated enthalpy change for this reaction? (Assume no changes in pressure or volume.)

A. 494 kJ/mol
B. 247 kJ/mol
C. –58 kJ/mol
D. –184 kJ/mol

Hydrogen H2 combines with chlorine Cl2 to form hydrochloric acid HCl H2 Cl2 2HCl Using the chart below what is the estimated enthalpy change for this reaction A class=

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Answer:

[tex]\large \boxed{\text{D. -184 kJ/mol}}[/tex]

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                       H₂  +  Cl₂ ⟶ 2HCl

Bonds:          1H-H   1Cl-Cl   2H-Cl

D/kJ·mol⁻¹:     436    242       431

[tex]\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 436 +242 - 2 \times 431\\&=& 678 - 862\\&=&\textbf{-184 kJ/mol}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-184 kJ/mol}}$}[/tex].

dcontreras0116

Answer:

Correct Answer: D. –184 kJ/mol

Explanation:

When pressure and volume are constant, the enthalpy change of a reaction can be determined by comparing the energy absorbed in breaking bonds to the energy released when bonds are formed.

In this reaction, one H–H bond and one Cl–Cl bond are broken, so the total energy absorbed is 436 + 242 = 678 kJ/mol. Then, two H–Cl bonds are formed, so the total energy released is 431 + 431 = 862 kJ/mol. The enthalpy change is equal to the energy absorbed minus energy released, so we have 678 – 862 = –184 kJ/mol. Because the enthalpy change is negative, the reaction is exothermic.

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