Answer:
4. The reaction is spontaneous at any temperatures if the standard entropy of reaction is positive.
Explanation:
Equilibrium constant decreases on increasing the temperature . So the reaction appears to be exothermic . In other words
ΔH is negative .
Δ G = ΔH - TΔS
If ΔS is positive , second term is negative . ΔH is negative so the RHS of the equation is negative at all temperatures . Hence Δ G of the reaction is negative . So reaction is spontaneous at any temperature.
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
Answer:
C. Graph C
Explanation:
We have a mixture of water and ice.
At 0 °C they are at equilibrium.
water-to-ice rate = ice-to-water rate
Next, we lower the temperature to -3 °C — just slightly below freezing.
The water will slowly turn to ice.
The water-to-ice rate will be slightly faster than the ice-to-water rate.
The purple bar will be slightly higher than the blue bar.
Graph C best represents the relative rates
A. is wrong. The ice-to-water rate is faster, so the water is melting. The temperature is slightly above freezing (say, 3 °C).
B. is wrong. The two rates are equal, so the temperature is 0 °C.
D. is wrong. The water-to-ice rate (freezing) is much greater than the ice-to-water rate, so the temperature is well below freezing( say, -10 °C).
