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Fudgin [204]
2 years ago
12

Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ

ine whether the reaction is first order or second order. determine whether the reaction is first order or second order. this is the first order reaction. this is the second order reaction. submitprevious answers correct part b what is the rate constant?
Chemistry
1 answer:
Whitepunk [10]2 years ago
7 0
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

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Mars2501 [29]

Answer: The reaction is not at equilibrium and will proceed to make more products to reach equilibrium.

Explanation:  

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium, while Q is the reaction  quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

H_2(g)+I_2(g)\rightarrow 2HI(g)

The expression for Q is written as:

Q=\frac{[HI]^2}{[H_2]^1[I_2]^1}

Q=\frac{[0.0890]^2}{[0.215]^1[0.498]^1}

Q=0.074

Given : K_{eq} = 54.8

Thus as Q, the reaction will shift towards the right i.e. towards the product side.

4 0
2 years ago
When 50 ml (50 g) of 1.00 m hcl at 22oc is added to 50 ml (50 g) of 1.00 m naoh at 22oc in a coffee cup calorimeter, the tempera
vitfil [10]

Answer:

\boxed{\text{2700 J}}

Explanation:

HCl + NaOH ⟶ NaCl + H₂O

There are two energy flows in this reaction.  

Heat of reaction + heat to warm water = 0  

           q₁             +              q₂                 = 0  

           q₁             +          mCΔT              = 0  

Data:

    m(HCl) = 50 g

m(NaOH) = 50 g

           T₁ = 22       °C

          T₂ = 28.87 °C

           C = 4.18 J·°C⁻¹g⁻¹

Calculations:

 m = 50 + 50 = 100 g

ΔT = 28.87 – 22 = 6.9 °C

 q₂ = 100 × 4.18 × 6.9 = 2900 J

q₁ + 2900 = 0

q₁ = -2900 J

The negative sign tells us that the reaction produced heat.

The reaction produced \boxed{\textbf{2900 J}}.

7 0
2 years ago
Convert 1.72 moles of magnesium carbonate to formula units
snow_tiger [21]
One mole any substance contains 6.022 ₓ 10²³ particles called Avogadro's Number.

The relation between moles and number of particles is given as,

                        # of particles  =  moles ₓ Avogadro's number

In our case the particles are formula units of MgCO₃. So, 1 mole of MgCO₃ contain 6.022 ₓ 10²³ formula units, then the number of formula units in 1.72 moles are calculated as,

              # of formula units  =  1.72 mol ₓ 6.022 ₓ 10²³ formula units / mol
            
             # of formula units   =  1.035 ₓ 10²⁴ Formula Units
8 0
2 years ago
One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer
sweet [91]

Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

How all the data are expressed with two significant figures, the volume must also be expressed with two.

7 0
2 years ago
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A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

<u>Explanation</u>:

  •  When dealing with dilution we will use the following equation:

                              M1 V1 = M2 V2

where,

                      M1 = initial concentration

                      V1 = initial volume

                      M2 = final concentration

                      V2 = final volume

  • By diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL, we get

                            M1 V1 = M2 V2

     20.0 mL \times    5.00 M = M2 \times 500.0 mL

                               M2 = (20.0 mL \times    5.00 M) / 500.0 mL

                              M2 =  0.200 M.

Hence A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.  

3 0
2 years ago
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