Answer:
This reaction is exothermic because the system shifted to the left on heating.
Explanation:
2NO₂ (g) ⇌ N₂O₄(g)
Reactant => NO₂ (dark brown in color)
Product => N₂O₄ (colorless)
From the question given above, we were told that when the reaction at equilibrium was moved from room temperature to a higher temperature, the mixture turned dark brown in color.
This simply means that the reaction does not like heat. Hence the reaction is exothermic reaction.
Also, we can see that when the temperature was increased, the reaction turned dark brown in color indicating that the increase in the temperature favors the backward reaction (i.e the equilibrium shift to the left) as NO₂ which is the reactant is dark brown in color. This again indicates that the reaction is exothermic because an increase in the temperature of an exothermic reaction will shift the equilibrium position to the left.
Therefore, we can conclude that:
The reaction is exothermic because the system shifted to the left on heating.
Answer:
Bonds are polar when one element in a compound is more electronegative than the other. This creates a dipole in the molecule where one end of the molecule is partially positive and one end is partially negative
Explanation:
Answer:
Percent yield = 84.5 %
Explanation:
Given data:
Mass of methanol = 229 g
Actual yield of water = 219 g
Percent yield of water = ?
Solution:
Chemical equation:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
Number of moles of methanol:
Number of moles = mass/ molar mass
Number of moles = 229 g/ 32 g/mol
Number of moles = 7.2 mol
Now we will compare the moles of water with methanol.
CH₃OH : H₂O
2 : 4
7.2 : 4/2×7.2 = 14.4 mol
Mass of water:
Mass = number of moles × molae mass
Mass = 14.4 mol × 18 g/mol
Mass = 259.2 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 219 g / 259.2 g × 100
Percent yield = 84.5 %
Answer:
1.346 v
Explanation:
1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:
(oxidation) 
→
E°=0.337 v
(reduction) 
→
E°=1.679 v
(overall) 
+8H^{+}_{(aq)}→
E°=1.342 v
2) Nernst Equation
Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:
![E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}](https://tex.z-dn.net/?f=E%3DE%5E%7B0%7D%20-%5Cfrac%7BRT%7D%7BnF%7DLn%5Cfrac%7B%5Bred%5D%7D%7B%5Box%5D%7D)
Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.
The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give
![E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346](https://tex.z-dn.net/?f=E%3D1.342%20-%5Cfrac%7B298.15%2A8.314%7D%7B6%2A96500%7DLn%5Cfrac%7B%5B.66%5D%7D%7B%5B1.69%5D%7D%3D1.346)
E=1.346
