Can you please tell me what the procedures are
<span>the formation of a gas
</span>
Answer:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, the rate is given as:
![rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}](https://tex.z-dn.net/?f=rate%3D-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%20%5BHBr%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B%5CDelta%20%5BBr_2%5D%7D%7B%5CDelta%20t%7D%20%3D%5Cfrac%7B%5CDelta%20%5BH_2%5D%7D%7B%5CDelta%20t%7D)
It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.
Therefore, the answer is:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
Best regards.
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!!
