Determine the rate law and the value of k for the following reaction using the data provided: 2n2o5(g) → 4 no2(g) + o2(g)0.2241.16 x 10-3 a. rate = 5.6 × 10-2 m-1s-1[n2o5]2 b. rate = 6.0 × 10-1 m-2s-1[n2o5]3 c. rate = 1.6 × 10-3 m1/2s-1[n2o5]1/2 d. rate = 1.7 × 10-2 m-1/2s-1[n2o5]3/2 e. rate = 5.2 × 10⁻3 s-1[n2o5]

The rate law and the value of k is Rate = 5.2 x 10^-3 s-1[N2O5]

The rate law or rate equation for a chemical reaction is an equation that links the reaction rate with concentrations or pressures of reactants and constant parameters (normally rate coefficients and partial reaction orders).

The correct answer between all the choices given is the last choice or letter E. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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