The enthalpy change for two reactions are given by the equations 2CrX(s)+1.5OX2X(g)⟶CrX2OX3X(s); ∆H1 = −1130 kJ ............(i) CX(s)+0.5OX2X(g)⟶COX(g); ∆H2 = −110 kJ .........(ii) - Chemistry

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MCQ

The enthalpy change for two reactions are given by the equations

\[\ce{2Cr_{(s)} + 1.5 O2_{(g)} -> Cr2O3_{(s)}}\];

∆H1 = −1130 kJ ............(i)

\[\ce{C_{(s)} + 0.5 O2_{(g)} -> CO_{(g)}}\];

∆H2 = −110 kJ .........(ii)

What is the enthalpy change, in kJ, for the following reaction?

\[\ce{3C_{(s)} + Cr2O3_{(s)} -> 2Cr_{(s)} + 3CO_{(g)}}\]

Options

  • −1460

  • 1460

  • −800

  • 800

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Solution

800

Explanation:

Reversing equation (i), multiplying equation (ii) by 3 and adding, we get,

∆H = −∆H1 + 3∆H2

= 1130 + 3 × (−110)

= 1130 − 330

= 800 kJ

Concept: Enthalpy (H)
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