From the standard enthalpies of formation values for some compounds (in kJ mol-1), predict which one of them will be most stable?

Question

From the standard enthalpies of formation values for some compounds (in kJ mol-1), predict which one of them will be most stable? (A) Propane (g)(Δ H°f=-103.8) (B) n-Butane (g)(Δ H°f=-124.7) (C) n-Butane (g)(Δ H°f=-167.2) (D) n-Octane (g)(Δ H°f=-208.4)

Tardigrade Admin · Accepted Answer

Greater the standard enthalpy of formation, the more stable will be the compound

Q. From the standard enthalpies of formation values for some compounds (in kJ mol^{-1}), predict which one of them will be most stable?

Propane $(g) (Δ H_{f}^{\circ} = - 103.8)$

n-Butane $(g) (Δ H_{f}^{\circ} = - 124.7)$

n-Butane $(g) (Δ H_{f}^{\circ} = - 167.2)$

n-Octane $(g) (Δ H_{f}^{\circ} = - 208.4)$

Greater the standard enthalpy of formation, the more stable will be the compound

Q. From the standard enthalpies of formation values for some compounds (in kJ mol^{-1}), predict which one of them will be most stable?

Propane (g)(ΔHf∘​=−103.8)

n-Butane (g)(ΔHf∘​=−124.7)

n-Butane (g)(ΔHf∘​=−167.2)

n-Octane (g)(ΔHf∘​=−208.4)

Greater the standard enthalpy of formation, the more stable will be the compound

Propane $(g) (Δ H_{f}^{\circ} = - 103.8)$

n-Butane $(g) (Δ H_{f}^{\circ} = - 124.7)$

n-Butane $(g) (Δ H_{f}^{\circ} = - 167.2)$

n-Octane $(g) (Δ H_{f}^{\circ} = - 208.4)$