Answer : The enthalpy change for the reaction is, 201.9 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The balanced reaction of
will be,
![\Delta H^o=?](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%3F)
The intermediate balanced chemical reaction will be,
(1)
![\Delta H_1=-890.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_1%3D-890.3kJ)
(2)
![\Delta H_2=-136.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_2%3D-136.3kJ)
(3)
![\Delta H_3=-571.6kJ](https://tex.z-dn.net/?f=%5CDelta%20H_3%3D-571.6kJ)
(4)
![\Delta H_4=-3120.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H_4%3D-3120.8kJ)
Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :
(1)
![\Delta H_1=2\times (-890.3kJ)=-1780.6kJ](https://tex.z-dn.net/?f=%5CDelta%20H_1%3D2%5Ctimes%20%28-890.3kJ%29%3D-1780.6kJ)
(2)
![\Delta H_2=-(-136.3kJ)=136.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_2%3D-%28-136.3kJ%29%3D136.3kJ)
(3)
![\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H_3%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%28-571.6kJ%29%3D285.8kJ)
(4)
![\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ](https://tex.z-dn.net/?f=%5CDelta%20H_4%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%28-3120.8kJ%29%3D1560.4kJ)
The expression for enthalpy of the reaction will be,
![\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5CDelta%20H_1%2B%5CDelta%20H_2%2B%5CDelta%20H_3%2B%5CDelta%20H_4)
![\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)](https://tex.z-dn.net/?f=%5CDelta%20H%3D%28-1780.6kJ%29%2B%28136.3kJ%29%2B%28285.8kJ%29%2B%281560.4kJ%29)
![\Delta H=201.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%3D201.9kJ)
Therefore, the enthalpy change for the reaction is, 201.9 kJ
Answer:
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Explanation:
Answer:
The sample of 50 mg decay to 10 mg in 377.7 time.
Explanation:
Given that:- The exponential decay model for the decay after t days as:-
Where,
is the concentration at time t
is the initial concentration
Given:
= 50 mg
= 10 mg
So,
![e^{-0.004261t}=\frac{1}{5}](https://tex.z-dn.net/?f=e%5E%7B-0.004261t%7D%3D%5Cfrac%7B1%7D%7B5%7D)
![t=\frac{\ln \left(5\right)}{0.004261}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cln%20%5Cleft%285%5Cright%29%7D%7B0.004261%7D)
![t=377.7](https://tex.z-dn.net/?f=t%3D377.7)
<u>The sample of 50 mg decay to 10 mg in 377.7 time.</u>