Hess's Law

Copper(II) nitrate decomposes on heating. The reaction is endothermic.

`2Cu(NO3)_2(s) -> 2CuO(s) + 4NO_2(g) + O_2(g)`

Draw an enthalpy cycle diagram to calculate the standard enthalpy change for this reaction, using enthalpy changes of formation.

When red phosphorus burns in oxygen the enthalpy change is –2967 kJ mol^–1. For white phosphorus the enthalpy change is –2984 kJ mol^–1. For both forms of phosphorus the reaction taking place is: 

`P_4(s) + 5O_2(g) -> P_4O_10(s)`

i. Use this information to calculate the enthalpy change for the transformation: P₄(white) -> P₄(red)

ii. Represent these changes on an enthalpy profile diagram.

i. 

ii.

Using Hess’s Law,  `ΔH""_r^theta - 2967 = -2984 => ΔH""_r^theta = -17 kJ mol^( -1)`

a. Define enthalpy change of solution.

b. Given the enthalpy changes ΔH₁ and ΔH₂ below, construct a Hess’s cycle that will enable you to find the enthalpy change, ΔH_r, for the reaction:

`MgCl_2(s) + 6H_2O(l) -> MgCl_(2).6H_2O(s) " "ΔH_r`

`MgCl_2(s) + aq -> MgCl_2(aq) " " ΔH_1`

`MgCl_(2).6H_2O(s) + aq -> MgCl_2(aq) " " ΔH_2`

a. The standard enthalpy change of solution is the enthalpy change when one mole of solute is dissolved in a solvent to form an infinitely dilute solution under standard conditions.

b. 

The table lists the relevant enthalpy of combustion data for the formation of propane from propyne.

Calculate the enthalpy change in kJ mol^-1 for the formation of propane from propyne. Show all working.

Enthalpy change of reaction, 

`ΔH""_r^theta = "total of" ΔH""_c^theta ("reactants") - "total of" ΔH""_c^theta ("products") = [(-1940) + (2xx-285.5)] - [(-2220)] = -291.6 kJ mol^(-1)`

Urea, CO(NH₂)₂ , is a naturally occurring substance which can be hydrolysed with water to form ammonia and carbon dioxide. The standard enthalpy changes of formation of water, urea, carbon dioxide and ammonia (in aqueous solution) are given below in Table

Construct a simple energy cycle for the hydrolysis of urea.

The standard enthalpy change of combustion, `ΔH""_c^theta` of 2-methylpropan-2-ol can be calculated using the data shown in Table below

i. State why oxygen does not have a value for `∆H""_f^theta`

ii. Construct the Hess' Law energy cycle for the combustion of 2- methylpropan-2-ol.

i. Because oxygen is an element and in its standard state

ii.

Construct a Hess’s Law cycle for the complete combustion of propanol.

Enthalpy changes that are difficult to measure directly can often be determined using Hess’ Law to construct an enthalpy cycle. Which enthalpy change is indicated by X in the enthalpy cycle shown?

A. +1 x Enthalpy of formation of water

B. -1 x Enthalpy of formation of water

C. +3 x Enthalpy of formation of water

D. -3 x Enthalpy of formation of water

The answer is C.

The standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements under standard conditions.

X shoes H₂O(l) being created from H₂(g) and O₂(g) however, 3 moles are produced, as such, x3 enthalpy of formation of water.

Hess’s Law can be used to calculate the enthalpy change for reactions that are difficult to measure experimentally, such as the conversion of graphite to diamond.

Which equation shows the correct application of Hess’s law to calculate the enthalpy change for the conversion of graphite to diamond?

A. `ΔH_r = ΔH_1 + ΔH_2`

B. `ΔH_r = ΔH_1 - ΔH_2`

C. `ΔH_r = ΔH_2 - ΔH_1`

D. `ΔH_r = ΔH_2 - ΔH_1`

The answer is B.

ΔH_r travels along the ΔH₁ arrow in same direction meaning that ΔH₁ is added while ΔH_r travels along the ΔH₂ arrow in opposite direction meaning that ΔH₂ is taken away. Thus, the final equation will be `ΔH_r=ΔH_1 - ΔH_2`

The combustion of ethanol (C₂H₅OH) is increasingly being used to fuel cars. 

The standard enthalpy change of formation of carbon dioxide is –382 kJ mol^-1.

The standard enthalpy change of formation of water is –275 kJ mol^-1.

The standard enthalpy change of formation of ethanol is −266kJ mol^-1.

What is the standard enthalpy change of combustion of ethanol?

A. -1367 kJ mol^-1

B. -1323 kJ mol^-1

C. -391 kJ mol^-1

D. -948 kJ mol^-1

The answer is B.

`C_2H_5OH(l) + 3O_2(g) -> 2CO_2(g) + 3H_2O(l)`

`ΔH_f = ΣΔH_f("products") -  ΣΔH_f("reactants") = [(2xx-382)+(3xx-275)] - [(-266)] = -1323 kJ mol^-1`