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lidiya [134]
3 years ago
8

If ammonia is manufactured at 356 k, is the reaction spontaneous, given that the enthalpy and entropy change for the reaction ar

e -93 kj/mol and -198 j/mol k, respectively?
Chemistry
1 answer:
Alchen [17]3 years ago
5 0
The reaction will be spontaneous if Gibb free energy is negative, according the following relation:  
G = H - T*S  
where G is Gibbs free energy, T is change of enthalpy , T is temperature and S is change of entropy  
In the case of ammonia:  
G = -93000 - 356*(-198) = -93000 + 70488 = -22512 j/mol  
As G < 0 then the reaction is spontaneous.
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Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6
MAXImum [283]

Answer:

the energy vacancies for formation in silver is \mathbf{Q_v = 3.069*10^{-4} \ J/atom}

Explanation:

Given that:

the equilibrium  number of vacancies at 800 °C

i.e T = 800°C     is  3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

N =  \dfrac{N_A* \rho _{Ag}}{A_{Ag}}

where ;

N_A = avogadro's number = 6.023*10^{23} \ atoms/mol

\rho _{Ag} = Density of silver = 9.5 g/cm³

A_{Ag} = Atomic weight of sliver = 107.9 g/mol

N =  \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}

N = 5.30 × 10²⁸ atoms/m³

However;

The equation for equilibrium number of vacancies can be represented by the equation:

N_v = N \ e^{^{-\dfrac{Q_v}{KT}}

From above; Considering the  natural logarithm on both sides; we have:

In \ N_v =In N - \dfrac{Q_v}{KT}

Making Q_v the subject of the formula; we have:

{Q_v =  - {KT}   In( \dfrac{ \ N_v }{ N})

where;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})

\mathbf{Q_v = 2.38 \ eV/atom}

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

Q_v =  (2.38 \ * 1.602176565 * 10^{-19} ) J/atom  }

\mathbf{Q_v = 3.069*10^{-4} \ J/atom}

Thus, the energy vacancies for formation in silver is \mathbf{Q_v = 3.069*10^{-4} \ J/atom}

8 0
3 years ago
The decomposition of ammonia is: 2 NH3(g) = N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni
hjlf

Answer:

A) = 4.7 × 10⁻⁴atm

Explanation:

Given that,

Kp = 1.5*10³ at 400°C

partial pressure pN2 = 0.10 atm

partial pressure pH2 = 0.15 atm

To determine:

Partial pressure pNH3 at equilibrium

The decomposition reaction is:-

2NH3(g) ↔N2(g) + 3H2(g)

Kp = [pH2]³[pN2]/[pNH3]²

pNH3 =√ [(pH2)³(pN2)/Kp]

pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm

K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm

= 4.7 × 10⁻⁴atm

4 0
3 years ago
Read 2 more answers
The reaction between zinc (Zn) and hydrogen chloride (HCl) produces zinc chloride (ZnCl), hydrogen (H2) gas, and heat. If more h
andrew-mc [135]

 If more   heat  is removed from the reaction the rate of reaction change as below to counter the action

The rate of the  <em>forward reaction increase</em> and produces more   <em>zinc chloride</em>


<u><em>      explanation</em></u>

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vekshin1
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