Pick the alkyl halide in the pair that reacts with sodium iodide in acetone at the faster rate. Select the single best answer. 1−Chlorohexane or Cyclohexyl chloride 1−Chlorohexane reacts at the faster rate. Cyclohexyl chloride reacts at the faster rate.

1-chlorohexane is faster

Explanation:

In this case, all we need to do is to know the function of the reagent in the reaction, in other words, we need to know if the reagent promoves a Sn1 or Sn2 reaction.

If the reagent promoves a Sn1 reaction, the faster rate will be the reactant with the most stable carbon, which follows the order 3° > 2° >> 1°.

If the reagent promoves a Sn2 reaction, the faster rate will be the reactant with the less stable carbon, in the order 1° > 2° >> 3°.

This is because, the Sn2 reaction is a bimolecular reaction that it's really fast and use only 1 step. It actually doesn't leave carbocations intermediate, that's why it's neccesary that the reactant has the less stability and least steric hindrance.

Taking this into account, the NaI in acetone is a reagent that always promoves either a Sn2 reaction or E2. As it's promoving the Sn2 reaction, let's see which is more stable between chlorohexane and cyclohexylchloride.

The chlorine in the cyclohexyl, it's on a secondary carbon ( CH2 - CH - Cl), while the chlorine in the chlorohexane it's on a primary carbon (CH2 - Cl); also the chlorohexane do not have steric hindrance, while the cyclohexyl has a little because of the cyclo, making the Sn2 reaction slower.

Therefore, the Chlorohexane would have the faster rate

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2