Question

At $\mathbf{10}\mathbf{°}\mathit{C}\mathbf{,}{\mathit{K}}_{\mathbf{e}\mathbf{q}}$for a reaction is 100 . At $\mathbf{30}\mathbf{°}\mathit{C}\mathbf{,}{\mathit{K}}_{\mathbf{e}\mathbf{q}}\mathbf{=}\mathbf{10}$Does enthalpy increase or decrease during the reaction?

Short answer

The enthalpy decreases during the reaction.

Step-by-step solution

Given data

Converting the temperature in Celsius to Kelvin,

$$\begin{gathered} {\text{T}}_{\text{1}}{\text{= 10}}^0\text{C} \\ \text{= 273.15 + 10} \\ {\text{T}}_{\text{1}}\text{= 283.15K} \end{gathered}$$

Similarly,

$$\begin{gathered} {\text{T}}_{\text{2}}{\text{= 30}}^0\text{C} \\ \text{= 273.15 + 30} \\ {\text{T}}_{\text{2}}\text{= 303.15K} \end{gathered}$$

Using the Van’t Hoff’s equation

$\text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{= -}\frac{∆H^0}{{\displaystyle \text{R}}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)$

Where,

${\text{K}}_{\text{2}}=$Equilibrium constant at$30°C$ .

${\text{K}}_{\text{1}}=$Equilibrium constant at $10°C$.

$\text{R =}$Gas constant${\text{= 8.3145JK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}$

$$\begin{gathered} {\text{T}}_{\text{2}}\text{= 303.15K} \\ {\text{T}}_{\text{1}}\text{= 283.15K} \end{gathered}$$

Calculating the enthalpy for the reaction

$$\begin{gathered} \text{ln}\left(\frac{\text{10}}{\text{100}}\right)\text{=-}\frac{∆H^0}{\text{8.314}}\left(\frac{\text{1}}{\text{303.15}}\text{-}\frac{\text{1}}{\text{283.15}}\right) \\ \text{- 2.3026 =}-\frac{∆H^0}{\text{8.314}}\left(\frac{\text{1}}{\text{303.15}}\text{-}\frac{\text{1}}{\text{283.15}}\right) \\ -∆H^0=\frac{\text{- 2.3026×8.3145}}{\text{- 0.00023}} \\ ∆H\text{= - 83.23899kJ} \end{gathered}$$

Therefore, the value of enthalpy is$∆H\text{= - 83.23899kJ}$

Hence, the value of enthalpy is negative which indicates that the enthalpy decreases during the reaction.