Question

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?

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

विकल्प

• −1460

• 1460

• −800

• 800

Answer 1

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