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Question: What is \({K_P}\) for the reaction \(2A(g) + 2B(g) \rightleftharpoons C(g)\), if \({K_C} = 34.8\) at...

What is KP{K_P} for the reaction 2A(g)+2B(g)⇌C(g)2A(g) + 2B(g) \rightleftharpoons C(g), if KC=34.8{K_C} = 34.8 at a temperature of 19oC{19^o}C.

Explanation

Solution

KP{K_P} the equilibrium constant used when equilibrium concentrations of reactants and products are given in terms of partial pressure whereas KC{K_C} is the equilibrium constant used when equilibrium concentrations of reactants and products are given in terms of molar concentration i.e., molarity.
Formula used-
KP=KC(RT)Δng{K_P} = {K_C}{(RT)^{\Delta {n_g}}}
Where, R is the universal gas constant expressed in unit of atm L mol−1K−1{\text{atm L mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}, T is the absolute temperature and Δng\Delta {n_g} is the difference between number of gaseous products and reactants.

Complete answer: As per question, the given data is as follows:
Equilibrium constant KC=34.8{K_C} = 34.8
Temperature T=19oCT = {19^o}C
Converting unit of temperature from degree Celsius to kelvin:
⇒T=19+273\Rightarrow T = 19 + 273
⇒T=292 K\Rightarrow T = 292\,K
The reaction is given as follows:
2A(g)+2B(g)⇌C(g)2A(g) + 2B(g) \rightleftharpoons C(g)
In the reaction, there is only one gaseous mole of product and four moles of gaseous reactants. Therefore, the value of Δng\Delta {n_g} will be as follows:
Δng=1−4\Delta {n_g} = 1 - 4
⇒Δng=−3\Rightarrow \Delta {n_g} = - 3
To calculate the value of KP{K_P} for the given reaction, substitute the values in the formula as follows:
KP=KC(RT)Δng{K_P} = {K_C}{(RT)^{\Delta {n_g}}}
⇒KP=34.8×(0.0821×292)−3\Rightarrow {K_P} = 34.8 \times {(0.0821 \times 292)^{ - 3}}
⇒KP=34.8×7.26×10−5\Rightarrow {K_P} = 34.8 \times 7.26 \times {10^{ - 5}}
⇒KP=252.6×10−5\Rightarrow {K_P} = 252.6 \times {10^{ - 5}}
⇒KP=2.53×10−3\Rightarrow {K_P} = 2.53 \times {10^{ - 3}}
Hence, the value of equilibrium constant KP{K_P} for the given reaction conditions is 2.53×10−32.53 \times {10^{ - 3}}.

Note:
It is important to note that the value of universal gas constant i.e., R is always considered in units of atm L mol−1K−1{\text{atm L mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}} for calculation of KP{K_P} because it is used when equilibrium concentrations are given in atmospheric pressure (atm). Also, some observations are made on the basis of ratio of KCKP\dfrac{{{K_C}}}{{{K_P}}} which are as follows:
i.Larger the value of KCKP\dfrac{{{K_C}}}{{{K_P}}}, the greater will be the percentage of product in the reaction.
ii.Lower the value of KCKP\dfrac{{{K_C}}}{{{K_P}}}, the greater will be the percentage of reactants in the reaction.