Respuesta :
Answer:
[tex]P_{H_2}=3.06atm[/tex]
Explanation:
Hello,
In this case, given the reaction:
[tex]4PH_3(g)\rightarrow P_4(g)+6H_2(g)[/tex]
We can represent the first-order reaction rate as:
[tex]r=\frac{dC_{PH_3}}{dt} =-kC_{PH_3}[/tex]
Whose integration turns out:
[tex]ln(\frac{C_{PH_3}}{C_{PH_3,}_0} )=-kt[/tex]
In such a way, we can compute the initial concentration by using the ideal gas equation and the given pressure and temperature:
[tex]C_{PH_3,}_0=\frac{P}{RT}=\frac{2.57atm}{0.082\frac{atm*L}{mol*K}*953K}=0.0329\frac{mol}{L}[/tex]
Next, by knowing the half-life, we can also compute the rate constant:
[tex]t_{1/2}=\frac{ln(2)}{k}\\ \\k=\frac{ln(2)}{t_{1/2}}=\frac{ln(2)}{35.0s}=0.0198s^{-1}[/tex]
Thus, we can find the concentration after 79.4 s have passed by solving for the concentration of phosphine from the integrated rate law:
[tex]C_{PH_3}=C_{PH_3,}_0}*exp(-kt)=0.0329\frac{mol}{L} *exp(-0.0198s^{-1}*79.4s)\\\\C_{PH_3}=6.83x10^{-3}\frac{mol}{L}[/tex]
Then, we can compute the consumed moles phosphine by using he container's volume and the initial and final concentrations:
[tex]n_{PH_3}=(0.0329-6.83x10^{-3})\frac{mol}{L}*4.60L=0.12molPH_3[/tex]
Hence, we compute the yielded moles of hydrogen by stoichiometry (4:6 molar ratio):
[tex]n_{H_2}=0.12molPH_3*\frac{6molH_2}{4molPH_3} =0.18molH_2[/tex]
Finally, we compute the partial pressure of hydrogen with the previously computed moles of hydrogen by using the ideal gas equation:
[tex]P_{H_2}=\frac{n_{H_2}RT}{V}=\frac{0.18mol*0.082\frac{atm*L}{mol*K}*953K}{4.6L}\\ \\P_{H_2}=3.06atm[/tex]
Best regards.
The study of chemicals and bonds is called chemistry. In a mixture of gases, when one gas molecule puts pressure on the other molecule is called partial pressure.
The correct answer to the question is 3.06atm.
Decomposition
- The process of breakdown of the complex molecule into the simple molecule is called decomposition. For example - the breakdown of carbohydrates into glucose.
According to the question, the solution is as follows:-
The formula used to solve the question is as follows:-
[tex]In\frac{C_{PH_3}}{C_{PH_3},0}[/tex].
The half rate of the equation is
[tex]t_\frac{1}{2} = \frac{In2}{k} \\\\k=\frac{In2}{t_\frac{1}{2} }[/tex], placed all the value to the question. after solving we will get 0.0198s
Then, we can compute the consumed mole's phosphine by using the container's volume and the initial and final concentrations:
[tex]nPh_3 =(0.0329-6.83*10^-3)*4.60=0.12 mole[/tex]
Hence, we compute the yielded moles of hydrogen by stoichiometry (4:6 molar ratio):
[tex]nH_2= 0.12moles\ of\ ph_3*\frac{6moleH_2}{4molePH_3}= 0.18moles[/tex]
Finally, we compute the partial pressure of hydrogen with the previously computed moles of hydrogen by using the ideal gas equation:
[tex]P_h_2 =\frac{nh_2Rt}{V} =\frac{0.18*0.82*953}{4.6L} =3.06atm\\3.06[/tex]
Hence, the correct answer is 3.06.
For more information about the law, refer to the link:-
https://brainly.com/question/13214440
