Areaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by o 3 ( g ) + no ( g ) ⟶ o 2 ( g ) + no 2 ( g ) the rate law for this reaction is rate of reaction = k [ o 3 ] [ no ] given that k = 2.55 × 10 6 m − 1 ⋅ s − 1 at a certain temperature, calculate the initial reaction rate when [ o 3 ] and [ no ] remain essentially constant at the values [ o 3 ] 0 = 7.34 × 10 − 6 m and [ no ] 0 = 6.99 × 10 − 5 m, owing to continuous production from separate sources.

1.31 × 10⁻¹⁵ M/s

Explanation:

Let's consider the following reaction.

O₃(g) + NO(g) ⟶ O₂(g) + NO₂(g)

The rate law is:

rate = 2.55 × 10⁻⁶ M⁻¹s⁻¹ [O₃].[NO]

where,

2.55 × 10⁻⁶ M⁻¹s⁻¹ is the rate constant (k)

[O₃] is the concentration of O₃ raised to the reaction order 1

[NO] is the concentration of NO raised to the reaction order 1

When [O₃] = 7.34 × 10⁻⁶ M and [NO] = 6.99 × 10⁻⁵ M, the rate of the reaction is:

rate = 2.55 × 10⁻⁶ M⁻¹s⁻¹ (7.34 × 10⁻⁶ M).(6.99 × 10⁻⁵ M) = 1.31 × 10⁻¹⁵ M/s

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