The hydrolysis of the sugar sucrose to the sugars glucose and fructose, C12H22011 + H20 - C6H12O6 + C H206
follows a first-order rate law for the disappearance of sucrose: rate = K[C12H22011] (The products of the reaction, glucose and
fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
In neutral solution, k = 2.1 x 10-11 s-1 at 27 °C

(b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65
x 10-7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the
concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
(c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

fun caring loving person

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

c. 4.54 x 10–12 j happens to be correct