Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.
<h3>What is the equilibrium constant?</h3>
In a reaction, we can judge using the value of the equilibrium constant weather or not the reaction moves on to completion. If the reaction moves up to completion, it the follows that the value of the equilibrium constant ought to be large.
On the other hand, when we have a case that the equilibrium constant is small and is not so large, then the reaction does not proceed essentially to completion.
Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.
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Answer:
<h3>hope its help u.................</h3>
Answer : The expression for reaction quotient will be :
(1) ![Q_c=\frac{[SO_2][HF]^4}{[SF_4]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BSO_2%5D%5BHF%5D%5E4%7D%7B%5BSF_4%5D%7D)
(2) ![Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}](https://tex.z-dn.net/?f=Q_c%3D%5Cfrac%7B%5BO_2%5D%5E2%5BXe%5D%7D%7B%5BXeF_2%5D%7D)
Explanation :
Reaction quotient 
: It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
(1) The given balanced chemical reaction is,
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :
(2) The given balanced chemical reaction is,
![2MoO_2(s)+XeF_2(g)\rightarrow 2MoF(l)+Xe(g)+2O_2(g)[/texIn this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :[tex]Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}](https://tex.z-dn.net/?f=2MoO_2%28s%29%2BXeF_2%28g%29%5Crightarrow%202MoF%28l%29%2BXe%28g%29%2B2O_2%28g%29%5B%2Ftex%3C%2Fp%3E%3Cp%3EIn%20this%20expression%2C%20only%20gaseous%20or%20aqueous%20states%20are%20includes%20and%20pure%20liquid%20or%20solid%20states%20are%20omitted.%20%20So%2C%20the%20expression%20for%20reaction%20quotient%20will%20be%20%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DQ_c%3D%5Cfrac%7B%5BO_2%5D%5E2%5BXe%5D%7D%7B%5BXeF_2%5D%7D)
Answer:
Explanation:
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<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
