Just as the depletion of stratospheric ozone today threatens life on earth today, its accumulation was one of the crucial processes that allowed life to develop in prehistoric times. you must have the correct sig figs.

3 o2 (g) > 2 o3 (g)
(a) express the rate of reaction in terms of the rate of disappearance of o2 and the rate of appearance of o3.
rate = (delta)[o2]/(delta)t and rate =)[o3]/(delta)t
(b) at a given instant, the rate of disappearance of o2 is 1.61 10-5 mol/ls. what is the rate of appearance of o3?
mol/ls

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt  

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

Explanation:

By definition, the reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:    

aX → bY (1)

where, a and b are the coefficients of de reactant X and product Y, respectively.        

(a) Based on the definition above, we can express the rate of reaction (2) as follows:      

3O₂(g) → 2O₃(g) (2)    

(3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:  

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

Have a nice day!

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