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Bromine and hydrogen react together to form hydrogen bromide - Scottish Highers Chemistry - Question 18 - 2017

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Bromine and hydrogen react together to form hydrogen bromide. $$H_2(g) + Br_2(g) \rightarrow 2HBr(g)$$ | Bonds broken | Bonds made | |--------------|------------| ... show full transcript

Worked Solution & Example Answer:Bromine and hydrogen react together to form hydrogen bromide - Scottish Highers Chemistry - Question 18 - 2017

Step 1

Bonds broken

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Answer

In the reaction, one mole of H–H and one mole of Br–Br bonds are broken. The total energy required to break these bonds is calculated as:

  • Energy required to break H–H: 436 kJ mol⁻¹
  • Energy required to break Br–Br: 194 kJ mol⁻¹

Thus, total energy for bonds broken is:

Ebroken=436+194=630 kJ mol−1E_{broken} = 436 + 194 = 630 \, \text{kJ mol}^{-1}

Step 2

Bonds made

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Answer

Two moles of H–Br bonds are formed in the reaction. The total energy released when these bonds are formed is:

  • Energy released for H–Br: 366 kJ mol⁻¹

Thus, total energy for bonds made is:

Emade=2×366=732 kJ mol−1E_{made} = 2 \times 366 = 732 \, \text{kJ mol}^{-1}

Step 3

Enthalpy change

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Answer

The enthalpy change \ (\Delta H) \ for the reaction can be calculated using the formula:

ΔH=Ebroken−Emade\Delta H = E_{broken} - E_{made}

Substituting the values we calculated:

ΔH=630−732=−102 kJ mol−1\Delta H = 630 - 732 = -102 \, \text{kJ mol}^{-1}

Therefore, the enthalpy change for the reaction is −102 kJ mol⁻¹, corresponding to option A.

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