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:
d. Sum of product enthalpies minus the sum of reactant enthalpies
Explanation:
The standard enthalpy change of a reaction (ΔH°rxn) can be calculated using the following expression:
ΔH°rxn = ∑n(products) × ΔH°f(products) - ∑n(reactants) × ΔH°f(reactants)
where,
ni are the moles of products and reactants
ΔH°f(i) are the standard enthalpies of formation of products and reactants
Explanation:
As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
..........(1)
..............(2)
The final reaction is as follows:
.............(3)
Therefore, adding (1) and (2) we get the final equation (3) and value of 
at 298 K will be as follows.
= 
+ 
= -314 kJ + (-80) kJ
= -394 kJ
Thus, we can conclude that 
at 298 K for the given process is -394 kJ.
I think your answer will be B. 273 k
Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
