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olya-2409 [2.1K]
2 years ago
12

9 The Haber process is a reversible reaction. N2(g) + 3H2(g) 2NH3(g) The reaction has a 30% yield of ammonia. Which volume of am

monia gas, NH3, measured at room temperature and pressure, is obtained by reacting 0.75 moles of hydrogen with excess nitrogen?
Chemistry
1 answer:
Illusion [34]2 years ago
3 0

Answer:  3.36 L of ammonia gas

Explanation:

The balanced chemical reaction is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)  

According to stoichiometry :

3 moles of H_2 produce = 2 moles of NH_3

Thus 0.75 moles of H_2 will producee=\frac{2}{3}\times 0.75=0.50moles  of NH_3

But as percent yield is 30 %, amount of ammonia produced = \frac{30}{100}\times 0.50moles=0.15moles

According to ideal gas equation:

PV=nRT

P = pressure  = 1 atm

V = Volume  = ?

n = number of moles = 0.15

R = gas constant =0.0821Latm/Kmol

T =temperature =273K

V=\frac{nRT}{P}

V=\frac{0.15\times 0.0820 L atm/K mol\times 273K}{1atm}=3.36L

Thus 3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.

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ω(solution) = 20% = 20% ÷ 100% = 0,2.
m(LiF) = ?
ω(solution) = m(LiF) ÷ m(solution).
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Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
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Answer : The value of K_{eq} is, 11.2

The value of \Delta G_{rxn} is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

K_{eq}  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

\Delta G^o=-RT\times \ln K_{eq}

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction  = ?

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

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Answer:

Explanation:

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Type the correct answer in the box. Express your answer to three significant figures.
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Answer:

The partial pressure of argon in the jar is 0.944 kilopascal.

Explanation:

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