The reaction depicted is:

Answer: Endothermic; ΔH 0 Products lie above reactants on the enthalpy axis, so ΔH is positive (endothermic).

Despite being endothermic, NH4NO3 dissolves spontaneously in water at 25 C. The thermodynamic explanation is:

Answer: the entropy of dissolution is positive and large enough that Tdelta S delta H delta G = delta H - Tdelta S < 0 requires delta S delta H / T. The disordering of the ionic lattice into solvated ions provides large positive delta S.

Based on the enthalpy diagram, what is the sign of ΔH for the reaction?

Answer: Negative; the reaction is exothermic Products lie below reactants on the enthalpy axis, so the system loses enthalpy and ΔH < 0 (exothermic).

AP Chemistry · Topic 6.2

Energy Diagrams Practice

Q: Based on the enthalpy diagram, what is the sign of ΔH for the reaction?

Answer: Negative; the reaction is exothermic Products lie below reactants on the enthalpy axis, so the system loses enthalpy and ΔH < 0 (exothermic).

Part of Thermodynamics.(ENE-2.B)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$The reaction depicted is:$
- A
  $Exothermic; ΔH < 0$
- B
  $Endothermic; ΔH > 0$
  check_circle
- C
  $Spontaneous regardless of T$
- D
  $At equilibrium$
Why
$Products lie above reactants on the enthalpy axis, so ΔH is positive (endothermic).$
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Sample 2difficulty 2/5
$A student dissolves 5.00 g of NH4NO3 (M = 80.04 g/mol) in 100.0 g of water in a coffee-cup calorimeter. Initial water temperature = 22.0 C. After dissolution and complete mixing, the minimum temperature is 18.6 C. Specific heat of solution = 4.18 J/(g*C). Total mass of solution = 105.0 g. Calorimeter heat capacity is negligible.$
$Despite being endothermic, NH4NO3 dissolves spontaneously in water at 25 C. The thermodynamic explanation is:$
- A
  $the solution releases heat to the surroundings$
- B
  $the entropy of dissolution is positive and large enough that T*delta S > delta H$
  check_circle
- C
  $the reaction is actually exothermic$
- D
  $the calorimeter is poorly insulated$
Why
$delta G = delta H - T*delta S < 0 requires delta S > delta H / T. The disordering of the ionic lattice into solvated ions provides large positive delta S.$
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Sample 3difficulty 2/5
$The enthalpy axis shows reactants high, products low.$
$Based on the enthalpy diagram, what is the sign of ΔH for the reaction?$
- A
  $Negative; the reaction is exothermic$
  check_circle
- B
  $Cannot be determined from the diagram$
- C
  $Zero; the reaction is at equilibrium$
- D
  $Positive; the reaction is endothermic$
Why
$Products lie below reactants on the enthalpy axis, so the system loses enthalpy and ΔH < 0 (exothermic).$
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Sample 4difficulty 3/5
$The diagram shows a reaction that is$
- A
  $Exothermic$
- B
  $Athermal (ΔH = 0)$
- C
  $Spontaneous at all T$
- D
  $Endothermic$
  check_circle
Why
$Products are higher in energy than reactants \to ΔH > 0 (endothermic).$
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Sample 5difficulty 3/5
$Using Hess's law, ΔH for A \to C is$
- A
  $+80 kJ$
- B
  $-140 kJ$
- C
  $+140 kJ$
- D
  $-80 kJ$
  check_circle
Why
$ΔH(A \to B) + ΔH(B \to C) = +30 - 110 = -80 kJ.$
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