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Jobisdone [24]
4 years ago
9

Is the haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at 25 ∘c? n

2(g)+3h2(g)→2nh3(g) δh∘=−92.2 kj; δs∘=−199 j/k?
Chemistry
1 answer:
Ostrovityanka [42]4 years ago
6 0
The  industrial  synthesis  of  ammonia   is  spontaneous.

 to  to  prove this  one  calculate  the  gibbs  free  energy(delta G)

=   delta  G  =  Delta  H  -T delta  s

delta H = -92.2  kj  in  joules  -92.2  x1000 =-92200j
delta  s  =  -199j/k
T=  25+273=  298k

delta  G  is  therefore=   -92200-298(-199)=-32898j
since  delta G  is  negative the  reaction  is  therefore  spontaneous
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Consider the reaction: 2 SO2(g)+O2(g)→2 SO3(g) If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mm
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Answer:

a. Oxygen is the limiting reagent. n_{SO_3}^{Theoretical}=8.096x10^{-4}mol SO_3

b. Y=58.9%

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

n_{SO_2}=\frac{PV}{RT}=\frac{0.0658atm*0.2855L}{0.082\frac{atm*L}{mol*K}*315K} =7.273x10^{-3}molSO_2 \\n_{O_2}=\frac{PV}{RT}=\frac{0.0658atm*0.1589L}{0.082\frac{atm*L}{mol*K}*315K} =4.048x10^{-4}molO_2

Afterwards, by considering the properly balanced chemical reaction:

2SO_2(g)+O_2(g)-->2SO_3

We compute the oxygen's moles that completely reacts with the previously computed 7.273x10^{-3} moles of SO_2 as follows:

7.273x10^{-3}molSO_2*\frac{1molO_2}{2molSO_2} =3.6365x10^{-3}molO_2

That result let us know that the oxygen is the limiting reagent since just 4.048x10^{-4} moles are available in comparison with the 3.6365x10^{-3} moles that completely would react with 7.273x10^{-3} moles of SO_2.

Now, to compute the theoretical yield of sulfur trioxide, we apply the following stoichiometric relationship:

n_{SO_3}^{Theoretical}=4.048x10^{-4}molO_2*\frac{2molSO_3}{1molO_2} =8.096x10^{-4}mol SO_3

b. Percent yield.

At first, we must compute the collected (real) moles of sulfur trioxide:

n_{SO_3}^{real}=\frac{PV}{RT}=\frac{0.0658atm*0.1872L}{0.082\frac{atm*L}{mol*K}*315K} =4.769x10^{-4}molSO_3

Finally, we compute the percent yield:

Y=\frac{n_{SO_3}^{real}}{n_{SO_3}^{Theoretical}} *100%

Y=\frac{4.769x10^{-4}mol SO_3}{8.096x10^{-4}mol SO_3} *100%

Y=58.9%

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