The answer is B . Brønsted-Lowry Acid/bases trade H+
To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the <span>∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.</span>
Answer: The value of the equilibrium constant Kc for this reaction is 3.72
Explanation:
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
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 
For the given chemical reaction:
The expression for is written as:
![K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E3%5Ctimes%20%5BH_2O%5D%5E1%7D%7B%5BHNO_3%5D%5E2%5Ctimes%20%5BNO%5D%5E1%7D)
Thus the value of the equilibrium constant Kc for this reaction is 3.72
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
Answer:
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