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This is the second of our many quizzes on Equilibrium constants. Click on this to access more such competitive quizzes on Physical Chemistry.

Equilibrium constants Quiz 2

chemistry

 In the Contact process for manufacturing sulphuric acid, the ______ pressure and _____ temperature favours forward reaction.

 The Le-Chatelier principle has the application in production of some chemicals which applies to their equilibrium constants where the pressure and the temperature are not the same for the reason of increasing the reaction rates.

 Contact process is the process used for the manufacture of sulphuric acid and the reaction is, $2S{{O}_{2}}+{{O}_{2}}2S{{O}_{3}}\,\,\,\,\left( \Delta H=-197kJ \right).$ Here, by looking at the value of enthalpy, the reaction is exothermic in nature and hence temperature used must be high for the economical process and the chemical equilibrium shifts to right according to Le-Chatelier principle and also lower pressure is employed to increase the reaction rates.

 The equilibrium constant of a reaction will be ___________ if the reaction is reversed.

 The reversal of the reaction is the conversion of products into reactant simultaneously and this in turn would inverse the equilibrium expression value.

 Reversible reaction is basically the reaction in which the reactants giving products and products giving reactions take place simultaneously. Thus, the equilibrium constant ‘K’ for this reaction will be reversed that is the reciprocal of the original one when the reaction is reversed and this is given by, ${{K'}}=\dfrac{1}{K}$ 

 What will be the equilibrium pressure if the dissociation constant ${{K}_{p}}$ of $N{{H}_{4}}HS(s)$ is $1.15\times {{10}^{-1}}$ at ${{25}^{0}}C$ ?

The dissociation of $N{{H}_{4}}HS(s)$ is the heterogeneous equilibrium reaction where the reactants and products are found in two or more phases. This particular reaction of solid produces gaseous products. 

 The equilibrium reaction of dissociation of $N{{H}_{4}}HS(s)$ is as shown below, $N{{H}_{4}}HS(s)N{{H}_{3}}(g)+{{H}_{2}}S(g).$ The dissociation constant can be written as, ${{K}_{p}}={{P}_{N{{H}_{3}}}}\times {{P}_{{{H}_{2}}S}} \\ \text{Since, } {{P}_{N{{H}_{3}}}}= {{P}_{{{H}_{2}}S}}$, the equation can be written as, ${{K}_{p}}=P_{N{{H}_{3}}}^{2}$ Taking the square root and substituting the given value, we have ${{P}_{N{{H}_{3}}}}=\sqrt{1.15\times {{10}^{-1}}}=3.39\times {{10}^{-1}}atm$. Now, equilibrium pressure will be= ${{P}_{N{{H}_{3}}}}+{{P}_{{{H}_{2}}S}}$. Substituting the values we have = $2\times 3.39\times {{10}^{-1}}=0.678atm$ 

 For a reaction whose standard enthalpy change is $-100 kJ$, what final temperature is needed to double the equilibrium constant from its value to $298K$?

The temperature dependence of the equilibrium constant for a reversible reaction is calculated using the formula $log\dfrac{{{K}_{2}}}{{{K}_{1}}}=\dfrac{\Delta {{H}^{0}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]$

 The Van’t Hoff factor for the temperature dependence of the equilibrium constant for a reversible reaction is, $log\dfrac{{{\left( {{K}_{p}} \right)}_{2}}}{{{\left( {{K}_{p}} \right)}_{1}}}=\dfrac{\Delta {{H}^{0}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]$ Since, the equilibrium constant is doubled, the above equation by substituting the necessary values will reduce to, $log\dfrac{2}{1}=\dfrac{-100000}{2.303\times 8.314}\left[ \dfrac{{{T}_{2}}-298}{298\times {{T}_{2}}} \right].$ Now, solving this for the temperature ${{T}_{2}},\\ 0.301=\dfrac{-100000}{19.09}\left[ \dfrac{{{T}_{2}}-298}{298\times {{T}_{2}}} \right].$ The approximate value of ${{T}_{2}}$ is $\approx 293K$ 

The equilibrium constant for the Haber process $\dfrac{3}{2}{{H}_{2}}+\dfrac{1}{2}{{N}_{2}}N{{H}_{3}}$ is $668$ at $300K$ and $6.04$ at $400K$. What is the average enthalpy of the reaction for the process in that temperature range?

 The Haber’s process of production of ammonia is exothermic in nature and hence the equilibrium constant decreases with the increase in temperature.

 The Van’t Hoff integrated equation to understand the effect of temperature on the equilibrium constant is given by, $log\dfrac{{{K}_{2}}}{{{K}_{1}}}=\dfrac{\Delta {{H}^{0}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right].$ By substituting the values according to the data given, we have $\log \dfrac{6.04}{668}=\dfrac{\Delta {{H}^{0}}}{2.303\times 8.314}\left[ \dfrac{400-300}{300\times 400} \right] \\ \Rightarrow \log \left( 3.042\times {{10}^{-3}} \right)=\dfrac{\Delta {{H}^{0}}}{19.09}\left( \dfrac{100}{120000} \right)$ Thus, when solved for $\Delta {{H}^{0}}$, we have $\Delta {{H}^{0}}=-2.044\times 22908=-46818\approx -47kJ/mol$ 