Question

The effect of temperature on equilibrium constant is expressed as $({{T}_{2}}>{{T}_{1}})$ $\log \dfrac{{{K}_{2}}}{{{K}_{1}}}=\dfrac{-\Delta H}{2.301}[\dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}}]$. For endothermic reaction, a false statement is:

a.) $[\dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}}]$ = positive
b.) $\Delta H = positive$
c.) $\log {{K}_{2}}>\log {{K}_{1}}$
d.) ${{K}_{2}}>{{K}_{1}}$

Answer 1

Hint: We know that the equilibrium constant is generally denoted as K, which is the ratio of the concentration of product and reactant when a reaction reaches an equilibrium state. The rate constant of a reaction is nearly doubled with a 10 degree rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by the Arrhenius equation.

Complete Solution:
It is given that,
$\log \dfrac{{{K}_{2}}}{{{K}_{1}}}=\dfrac{-\Delta H}{2.301}[\dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}}]$
A. An endothermic reaction is a reaction that absorbs heat from the surroundings. So, the surrounding temperature is decreased by the endothermic reaction. For an endothermic process, heat is required. Hence, $\Delta H=positive$

B. Also, $({{T}_{2}}>{{T}_{1}})$, it means, $[\dfrac{1}{{{T}_{2}}}<\dfrac{1}{{{T}_{1}}}]$, so A is negative.
 It simply means ${{T}_{2}}$ is at a higher temperature than ${{T}_{1}}$
So,
$[\dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}}]$
By solving this, we get
\[[\dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}}]<0\]
Hence, ${{T}_{1}}-{{T}_{2}}<0$

C. As we know,
$\log \dfrac{{{K}_{2}}}{{{K}_{1}}}=\dfrac{-\Delta H}{2.301}[\dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}}]$
So, $\log \dfrac{{{K}_{2}}}{{{K}_{1}}}>0$
Therefore, $\log {{K}_{2}}>{{K}_{1}}$
Hence, the false statement is $[\dfrac{1}{{{T}_{2}}}-\dfrac{1}{{{T}_{1}}}]$=positive

So, the correct answer is “Option A”.

Additional Information:
Characteristics of Equilibrium constant ${{K}_{C}}$:
1) For a particular reaction at a given temperature, it has a constant value.
2) Its value is independent of the initial concentration of reactants and products.
3) Its value is dependent on the nature of reactants and temperature, but independent of the presence of a catalyst.
4) Its value tells the extent to which reaction proceeds in the forward and reverse direction.
5) For a reversible reaction, the equilibrium constant for the forward reaction is equal to the inverse of the equilibrium constant of backward reaction.
6) Equilibrium constants are achieved by using the relation between equilibrium concentrations of products and reactants.

Note: The possibility to make a mistake is that when the temperature is increased, the endothermic reaction is favored, not the exothermic reaction.