The reaction 2NO(g) + O2(g) 2NO2(g) is second order in NO and first order in O2. When [NO] = 0.8 M and [O2] = 3.7 M, the observed rate of the reaction is 0.00022022 M/s. (a) What is the value of the rate constant? (d) What is the rate of reaction when [NO] = 0.1 M and [O2] = 0.47 M?

a)  

b)

Explanation:

Hello,

In this case, given that the reaction is second order in NO and first order in O2, the rate law is written as follows:

In such a way, for a rate law of 0.00022022 M/s and the concentrations of NO and O2 0.8M and 3.7M respectively, the rate constant is:

Thus, for the new concentrations of NO and O2 0.1M and 0.47M respectively, the rate is:

Regards.

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).

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