<u>Answer:</u>
<u>For A:</u> The expression for
is given below.
<u>For B:</u> The value of
at 25°C is 0.0185512
<u>For C:</u> The value of
at 65°C is 0.2887886
<u>For D:</u> The reaction is endothermic in nature.
<u>Explanation:</u>
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as ![K_{eq}](https://tex.z-dn.net/?f=K_%7Beq%7D)
For the given chemical reaction:
![[Co(H_2O)_6]^{2+}(aq.)+4Cl^-(aq.)\rightleftharpoons [CoCl_4]^{2-}(aq.)+6H_2O(l)](https://tex.z-dn.net/?f=%5BCo%28H_2O%29_6%5D%5E%7B2%2B%7D%28aq.%29%2B4Cl%5E-%28aq.%29%5Crightleftharpoons%20%5BCoCl_4%5D%5E%7B2-%7D%28aq.%29%2B6H_2O%28l%29)
The expression of
for above equation without the concentration of liquid water is:
......(1)
The expression is written above.
We are given:
![[CoCl_4]^{2-}=0.0334612M](https://tex.z-dn.net/?f=%5BCoCl_4%5D%5E%7B2-%7D%3D0.0334612M)
![[Co(H_2O)_6]^{2+}=0.966539M](https://tex.z-dn.net/?f=%5BCo%28H_2O%29_6%5D%5E%7B2%2B%7D%3D0.966539M)
![[Cl^-]=1.86616M](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D1.86616M)
Putting values in equation 1, we get:
![K_{eq}=\frac{0.0334612}{0.966539\times 1.86616}=0.0185512](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B0.0334612%7D%7B0.966539%5Ctimes%201.86616%7D%3D0.0185512)
Hence, the value of
at 25°C is 0.0185512
We are given:
![[CoCl_4]^{2-}=0.234625M](https://tex.z-dn.net/?f=%5BCoCl_4%5D%5E%7B2-%7D%3D0.234625M)
![[Co(H_2O)_6]^{2+}=0.765375M](https://tex.z-dn.net/?f=%5BCo%28H_2O%29_6%5D%5E%7B2%2B%7D%3D0.765375M)
![[Cl^-]=1.06150M](https://tex.z-dn.net/?f=%5BCl%5E-%5D%3D1.06150M)
Putting values in equation 1, we get:
![K_{eq}=\frac{0.234625}{0.765375\times 1.06150}=0.2887886](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B0.234625%7D%7B0.765375%5Ctimes%201.06150%7D%3D0.2887886)
Hence, the value of
at 65°C is 0.2887886
For Endothermic reactions,
, which is positive
For Exothermic reactions,
, which is negative
To calculate
of the reaction, we use Van't Hoff's equation, which is:
![\ln(\frac{K_{65^oC}}{K_{25^oC}})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B65%5EoC%7D%7D%7BK_%7B25%5EoC%7D%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 65°C = 0.2887886
= equilibrium constant at 25°C = 0.0185512
= Enthalpy change of the reaction = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![25^oC=[25+2730]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B2730%5DK%3D298K)
= final temperature = ![65^oC=[65+2730]K=338K](https://tex.z-dn.net/?f=65%5EoC%3D%5B65%2B2730%5DK%3D338K)
Putting values in above equation, we get:
![\ln(\frac{0.2887886}{0.0185512})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{298}-\frac{1}{338}]\\\\\Delta H=57471.26J/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B0.2887886%7D%7B0.0185512%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B298%7D-%5Cfrac%7B1%7D%7B338%7D%5D%5C%5C%5C%5C%5CDelta%20H%3D57471.26J%2Fmol)
As, the calculated value of
. Thus, the reaction is endothermic in nature.