## How much Van’t Hoff equation – effect on temperature?

**Van’t Hoff equation**–equilibrium of a chemical reaction is constant at a given temperature. The equilibrium constant, Kp values can be changed with the change of temperature for an endothermic and exothermic reaction.

This will be evident from the study of Kp values at different temperatures of a chemical reaction.

N_{2} + O_{2} ⇆ 2NO

Temperature | Kp × 10^{4} |

2000° K | 4.08 |

2200° K | 11.00 |

2400° K | 25.10 |

2600° K | 50.30 |

The quantitative relation, known Van’t Hoff equation connecting chemical equilibrium and temperature can be derived thermodynamically starting from Gibbs – Helmholtz equation.

The Gibbs – Helmholtz equation

Zero superscripts are indicating stranded values.

Van’t Hoff isotherm

– RT lnK_{P} = ΔG^{0}

or, – R lnK_{P} = ΔG^{0}/T

Differentiating with respect to temperature at constant pressure

Comparing the above two-equation

This is the differential form of Van’t Hoff equation.

#### Derive van’t Hoff general equation

The greater the value of standard enthalpy of a reaction, the faster the chemical reaction reaching an equilibrium point.

Separating the variables and integrating

ΔH^{0} independent of temperature.

where C = integrating constant.

The integration constant can be evaluated and identified the value of ΔS^{0}/R, using the relation

ΔG^{0} = ΔH^{0} – TΔS^{0}

∴ Van’t hoff equation

Problem

The standard free energy of chemical reaction 2A + B ⇆ 2C at 500 K = 2 KJ mol^{-1}. Calculate Kp at 500 K for the chemical reaction A + ½B ⇆ C.

Solution

Standard free energy at 500 K for the chemical reaction

= 2 kJ mol^{-1}/2

= 1 kJ mol^{-1}

ΔG^{0} = – RT lnK_{P}

∴ 1 = – 8.31 × 10^{-3} × 500 × lnK_{P}

or, lnK_{P} = 1/(8.31 × 0.5)

= 0.2406

∴ K_{P} = 1.27

### Enthalpy of reaction endothermic or exothermic

Heat absorption and emission in a chemical reaction can be studied from the following analysis.

We may study the following possibilities

#### Endothermic reaction, ΔH^{0} = positive

If a specific heat absorbed, the forward direction of the chemical reaction is the endothermic direction and the reverse one is the exothermic direction. The dissociation of hydrogen iodide into hydrogen and Iodine is an example of this type of chemical reaction.

HI ⇆ H_{2} + I_{2} |
ΔH^{0} = (+) ve |

#### An exothermic reaction, ΔH^{0} = negative

If heat emission, the forward direction of the chemical reaction is the exothermic direction and the reverse one is the endothermic direction.

The formation of ammonia from hydrogen and nitrogen is an example of this type of chemical reaction.

N_{2} + H_{2} ⇆ 2NH_{3} |
ΔH^{0} = (-) ve |

### The endothermic and exothermic reaction

Van’t Hoff equation drowns the relation between equilibrium constant and **endothermic and exothermic reactions**.

- For an endothermic reaction, ΔH
^{0}〉 0 and the right-hand side of the equation positive. This leads to the fact that lnKp increases with increasing temperature. - For an
**exothermic reaction**, ΔH^{0}〈 0 and the right-hand side of the equation negative. This leads to the fact that lnKp decreases with increasing temperature.

#### Heat change of a chemical reaction in two temperature

For the ideal system, the heat change or thermodynamic enthalpy change is not a function of pressure.

^{0}= ΔH

However, the entropy of an ideal gas depends strongly on pressure and ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS^{0} and ΔG^{0}.

The integrated form of the Van’t Hoff equation at two temperature

where K_{P1} and K_{P2} are the equilibrium constants of the reaction at two different temperatures T_{1} and T_{2} respectively.

Determination of K_{P1} and K_{P2} at two temperatures helps to calculate the value of change of enthalpy of the chemical reaction.

The above relation called Van’t Hoff reaction isobar since pressure remains constant during the change of temperature.

Problem

Show that the equilibrium point for any chemical reaction given by ΔG = 0.

Solution

Van’t Hoff reaction isotherm

ΔG = – RT lnKa + RT lnQa.

When the reaction attains equilibrium point,

Qa = Ka.

∴ ΔG = 0

### Assumptions from Van’t Hoff equation

- The reacting system of the chemical reaction behaves ideally.
- ΔH has taken independent of temperature for a small range of temperature change.

Due to the assumption involved ΔH and ΔU do not produce the precise value of these reactions. As ΔG independent of pressure for ideal system ΔH and ΔU also independent of pressure.

ΔG^{0} = – RT lnK_{P}

Problem

Use Gibbs – Helmholtz equation to derive the Van’t Hoff reaction isochore. What condition do you expect a linear relationship between log k and 1/T?

Solution

K_{P} = K_{C} (RT)^{Δγ}

lnK_{P} = lnK_{C} + ΔγlnR + ΔγlnT

Differentiating with respect to T,

ΔU^{0} = standers heat of reaction at constant volume. This is the differential form of Van’t Hoff reaction isochore.

The integrated form of the equation at two temperature

For ideal system ΔU^{0} = ΔU. Since the reaction occurs at constant volume and the equation called Van’t Hoff reaction isochore.

#### The heat of a reaction and Le-Chatelier principal

Van’t Hoff equations give a quantitative expression of the Le-Chatelier principle.

- Endothermic reaction, ΔH〉0, an increase in temperature increases the value of lnKp of the chemical reactions.
- An exothermic reaction, ΔHㄑ 0, with rising in temperature, lnKp decreased.

Change of Kp provides the calculation of the quantitative change of equilibrium point yield of products.

According to the Le – Chatelier Principle, whenever stress placed on any system in a state of the equilibrium point, the system always reacts in a direction to reduce the applied stress.

- The temperature increased in a chemical reaction, the system at equilibrium point will try to move in a direction in which heat absorbed, which is endothermic reaction favors.
- When the temperature decreased in a chemical reaction, the system at equilibrium point will try to move in a direction in which heat emitted, which is exothermic reaction favors.