Answer:
Explanation:
These properties are governed by intermolecular forces. The most important part here is the oxygen interaction with the surroundings. For temperature the decreasing order is pentanol, pentanal and pentane. For viscocity: pentanol, pentanal and pentane. For surface tension: pentanol, pentanal and pentane.
This order, as said before, is due to the interaction of oxygen with the surroundings, within the intermolecular forces we can find van der waals forces and hydrogen bonds, it is also know that H-bonds are stronger than van der waals forces so then that is why we have this type of interactions.
Heat always flows from hot to cold.
Answer: 1) ![2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)](https://tex.z-dn.net/?f=2H_2%28g%29%2BS_2%28g%29%5Crightleftharpoons%202H_2S%28g%29)
Equilibrium constant is defined as the ratio of the product of concentration of products to the product of concentration of reactants each term raised to their stochiometric coefficients.
![K_{eq}=\frac{[H_2S]^2}{H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)
where [] = concentration in Molarity=![\frac{moles}{\text {Volume in L}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmoles%7D%7B%5Ctext%20%7BVolume%20in%20L%7D%7D)
Thus ![[H_2S]=\frac{68.5}{1.0}=68.5M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B68.5%7D%7B1.0%7D%3D68.5M)
![[H_2]=\frac{0.50}{1.0}=0.50M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B0.50%7D%7B1.0%7D%3D0.50M)
![[S_2]=\frac{0.020}{1.0}=0.020M](https://tex.z-dn.net/?f=%5BS_2%5D%3D%5Cfrac%7B0.020%7D%7B1.0%7D%3D0.020M)
![K_{eq}=\frac{[68.5]^2}{0.50]^2\times [0.020]}=938450](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5B68.5%5D%5E2%7D%7B0.50%5D%5E2%5Ctimes%20%5B0.020%5D%7D%3D938450)
As the value of K is greater than 1, the reaction is product favored.
2) ![N_2O_4(g)\rightleftharpoons 2NO_2(g)](https://tex.z-dn.net/?f=N_2O_4%28g%29%5Crightleftharpoons%202NO_2%28g%29)
![K_{eq}=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![K_{eq}=\frac{[0.500]^2}{[0.0250]}=10](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5B0.500%5D%5E2%7D%7B%5B0.0250%5D%7D%3D10)
3) ![N_2+3H_2\rightleftharpoons 2NH_3](https://tex.z-dn.net/?f=N_2%2B3H_2%5Crightleftharpoons%202NH_3)
![K_{eq}=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
4) Reactions which do not continue to completion are called equilibrium reactions as the rate of forward reaction is equal to the rate of backward direction.
Answer:
Ionisation energy increases along a period but decreases down a group
Explanation:
As you go along the period, the amount of protons in the nuclear increases. This causes nuclear attraction to increase meaning the attraction between the protons and the electrons is higher. This causes the electrons to be held more tightly to the nucleus meaning they are harder to remove. As well as that, the shielding stays the same along a period since the electrons are being added to the same shell, just different orbitals. Overall this makes it harder to remove an electron as you go along a period causing the ionisation energy to increase along a period.
On the other hand when you go down a group, the atomic radius increases. This is because the electrons are added to new shells which are further away from the nucleus, this also causes shielding to increase as there are more shells. This means the outer electrons being removed are held less tightly by the protons in the nucleus meaning they are easier to remove. This means that the ionisation energy decrease down a group.