Ozone, o3(g), is a form of elemental oxygen that is important in the absorption of ultraviolet radiation in the stratosphere. it decomposes to o2(g) at room temperature and pressure according to the following reaction: (a) what is the enthalpy change for this reaction per mole of o3(g)? (b) which has the higher enthalpy under these conditions,2 o3(g) or 3 o2(g)?

-142.3 per mole, b) 2 O₃ (g)

Explanation:

2O₃ (g) → 3O₂ (g)   ΔH = -284.6 KJ

According to the equation 2 mole of ozone yields 3 mole of oxygen gas and the enthalpy change = -284.6 KJ

a) enthalpy change per mole = -284.6 / 2 mole = -142.3 KJ per mole

b) The reaction is exothermic since the enthalpy change is negative meaning that the heat of reactant is greater than heat of product. If the heat of product is greater than heat of reactant meaning enthalpy change is positive; the reaction is endothermic. 2 O₃ (g) has higher enthalpy  compared to 3 O₂ (g) since the enthalpy change is negative

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