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Hoochie [10]
3 years ago
9

The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be

at 298oK? What will the half-life be at both temperatures?
Chemistry
1 answer:
Sedbober [7]3 years ago
3 0

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature T_1 = 1001 K

Final temperature T_2 = 298 K

Applying the equation of Arrhenius theory.

In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})

where ;

R gas constant = 8.314  J/K/mol

In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})

In \dfrac{k_2}{0.380 }= -70.8655

\dfrac{k_2}{0.380 }= e^{-70.8655}

\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}

{k_2}= 1.67303256 \times 10^{-31} \times {0.380 }

{k_2}= 6.3575 \times 10^{-32}  /M .sec

Half life:

At 1001 K.

t_{1/2} = \dfrac{In_2}{k_1}

t_{1/2} = \dfrac{0.693}{0.38}

t_{1/2} = 1.82368 secc

At 298 K:

t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}

t_{1/2} =1.09 \times 10^{31} \ sec

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kodGreya [7K]
Equation is as follow,
   
<span>                               CaCN</span>₂<span> + 3 H</span>₂<span>O   →   CaCO</span>₂<span> + 2 NH</span>₃

According to this equation, 
When,
100 g CaCO₃ (1 mole) is produced when  =  34 g NH₃ (2 moles) is produced
So,
187 g CaCO₃ will be produced then  =  X g of NH₃ will produce

Solving for X,
                                 X  =  (187 g × 34 g) ÷ 100 g

                                 X  =  63.58 g of NH₃ will be produced

Result:
          Option-2 is correct answer.
5 0
4 years ago
0.5 moles of sodium chloride is dissolved to make 0.05litres of solution. find the molar concentration
Gala2k [10]
M = n / V

M = 0.5 / 0.05

M = 10 mol/L-1

Hope this helps!
7 0
3 years ago
A pencil sharpener has a handle with a radius of 1.5 cm. The sharpening mechanism has a radius of 0.5 cm. What is the IMA of the
vivado [14]
The answer is A I believe
7 0
3 years ago
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What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3
snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

ΔG°=-16.7 kJ/mol

Required

the free energy change

Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol

8 0
3 years ago
How many litres of fluorine gas at stp can be collected from the decomposition of 90.7 g of AuF3
7nadin3 [17]

Answer: 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of AuF_3

Explanation:

The balanced decomposition reaction is shown as

2AuF_3\rightarrow 2Au+3F_2

moles of AuF_3=\frac{\text {given mass}}{\text {Molar mass}}=\frac{90.7g}{254g/mol}=0.36moles

According to stoichiometry:

2 moles of AuF_3 gives = 3 moles of flourine gas

Thus 0.36 moles of AuF_3 gives = \frac{3}{2}\times 0.36=0.54moles of flourine gas

Using ideal gas equation :

PV=nRT

P = pressure of gas = 1 atm ( at STP)

V = Volume of gas = ?

n = moles of gas = 0.54

R = gas constant = 0.0821 L atm/Kmol

T = temperature = 273 K ( at STP)

Putting the values we get :

1atm\times V=0.54mol\times 0.0821Latm/Kmol\times 273K

V=12L

Thus 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of AuF_3

6 0
3 years ago
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