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Question: The rate of the reaction \[{H_2} + {I_2} \leftrightarrow 2HI\] is given by Rate=\[1.7 \times {10^...

The rate of the reaction H2+I2↔2HI{H_2} + {I_2} \leftrightarrow 2HI is given by
Rate=1.7 \times {10^{ - 19}}$$$$\left[ {{H_2}} \right]\left[ {{I_2}} \right] at 25∘C25^\circ C
The rate of decomposition of gaseous HIHI to H2{H_2} and I2{I_2} is given by
Rate=2.4 \times {10^{ - 21}}$$$${\left[ {HI} \right]^2} at 25∘C25^\circ C
Calculate the equilibrium constant for the formation of HIHI to H2{H_2} and I2{I_2} at 25∘C25^\circ C?

Explanation

Solution

Firstly find equilibrium constant then, do the calculation by calculating the each step of finding the constant very carefully. Then finally put all the values to find the equilibrium constant.
kc=[HI]2[H2][I2]{k_c} = \frac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}

Complete step-by-step answer: A reversible reaction is a reaction where the reactants and the products react together to give back the reactants.
For the reversible reaction,
H2+I2↔2HI{H_2} + {I_2} \leftrightarrow 2HI
Talking then about the equilibrium constant we can say that the equilibrium constant of a chemical reaction is the value of the reactant quotient at chemical equilibrium then a state is reached by the dynamic chemical system after the sufficient time has passed at which its composition has no measurable tendency towards the further change or in common language we can say that the equilibrium is reached. According to the question the equilibrium constant is given by:
Equilibrium constant,
kc=[HI]2[H2][I2]{k_c} = \frac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}
At equilibrium,
Now, [H2][I2]\left[ {{H_2}} \right]\left[ {{I_2}} \right]
[HI]2{\left[ {HI} \right]^2}
1.7 \times {10^{ - 19}}$$$$\left[ {{H_2}} \right]\left[ {{I_2}} \right]
As given in the question.
=2.4×10−212.4 \times {10^{ - 21}}
Putting the values in the equation and solving it to get the value of the constant.
1.7×10192.4×10−21=[HI]2[H2][I2]\frac{{1.7 \times {{10}^{19}}}}{{2.4 \times {{10}^{ - 21}}}} = \frac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}
=Kc=70.83= {K_c} = 70.83

Note: Equilibrium constant should be determined carefully. Further calculations should be carried out very carefully.