Answer:
Explanation:
The chemical equation for the reaction is :
![C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}](https://tex.z-dn.net/?f=C_6H_%7B12%7DO_6_%7B%28s%29%7D%20%5Cto%202C_2H_5OH%20_%7B%28l%29%7D%20%2B%202CO_%7B2%28g%29%7D)
The standard enthalpy of formation
of the above equation is as follows:
= -1274.4 kJ/mol
= -277.7 kJ/mol
= -393.5 kJ/mol
![\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}](https://tex.z-dn.net/?f=%5CDelta%20H%20%20%5E0_%7Brxn%20%7D%3D%20%5Csum%20n_p%20%5CDelta%20H%20%20%5E0_%7Bf%2Cp%7D%20-%20%5Csum%20n_r%20%5CDelta%20H%20%20%5E0_%7Bf%2Cr%7D)
where ;
= stochiometric coefficients of products
stochiometric coefficients of reactants
= formation standard enthalpy of products
= formation standard enthalpy of reactants
![\Delta H ^0_{rxn }= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20H%20%20%5E0_%7Brxn%20%7D%3D%20%282%20%5C%20mol%2A%20-2777.7%20%5C%20kJ%2Fmol%20%2B%202%20%5C%20mol%20%2A%20-%20393.5%20%5C%20kJ%2Fmol%29%20-%20%281%5C%20mol%20%2A%28-1274.4%20%5C%20kJ%2Fmol%29)
![\Delta H ^0_{rxn }= -68 \ kJ](https://tex.z-dn.net/?f=%5CDelta%20H%20%20%5E0_%7Brxn%20%7D%3D%20-68%20%5C%20kJ)
For
;
The standard enthalpy of formation of
of the reactant and the products are :
![\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0%20_%7Bf%20%5C%20C_6H_%7B12%7DO_6%7D%20%3D%20212%20%5C%20%5C%20%20J%2FKmol)
![\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0%20_%7Bf%20%5C%20C_2H_5O_H%7D%20%3D%20160.7%20%5C%20%5C%20J%2FKmol)
![\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0%20_%7Bf%20%5C%20CO_2%7D%20%3D%20213.63%20%5C%20%5C%20J%2FKmol)
The
is as follows:
![\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D%20%5Csum%20n_p%20%5CDelta%20S%5E0_%7Bf.p%7D%20-%20%5Csum%20n_r%20%20%5CDelta%20S%5E0_%7Bf.r%7D)
![\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D%20%282%20%5C%20mol%20%2A160.7%20%5C%20%5C%20J%2FKmol%20%2B%202%20%5C%20mol%20%2A213.63%20%5C%20%5C%20J%2FKmol%29%20-%281%2A212%20J%2FKmol%29)
(to kJ/K)
![\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D536%20.%207%20%5C%20J%2FK%20%2A%20%5Cfrac%7B1%20%5C%20kJ%7D%7B1000%20%5C%20J%7D)
![\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}](https://tex.z-dn.net/?f=%5CDelta%20S%20%5E0_%7Brxn%7D%20%3D0.5367%20%5Cfrac%7BkJ%7D%7BK%7D)
Given that;
at T = 25°C = ( 25 + 273) K = 298 K
![\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}](https://tex.z-dn.net/?f=%5CDelta%20G%5E0%20_%7Brxn%7D%20%3D%20%5CDelta%20H%5E0_%7Brxn%7D%20-%20T%20%5CDelta%20S%5E0_%7Brxn%7D)
![\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}](https://tex.z-dn.net/?f=%5CDelta%20G%5E0%20_%7Brxn%7D%20%3D%20%20-68%20%5C%20kJ%20-%20298%20%2A%200.5367%20%5Cfrac%7BkJ%7D%7BK%7D)
![\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ](https://tex.z-dn.net/?f=%5CDelta%20G%5E0%20_%7Brxn%7D%20%3D%20%20-227.%209%20%5C%20%5C%20%5C%20kJ)
As
is negative; the reaction is spontaneous
= negative
= positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction
is dependent on temperature.
Answer:An isotonic solution refers to two solutions having the same osmotic pressure across a semipermeable membrane. This state allows for the free movement of water across the membrane without changing the concentration of solutes on either side. i hope this helps :) .
Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
Explanation:
Answer
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The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.
Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0
2× ( Oxidation number of Al) +3(−2)=0
2× ( oxidation number of Al) +6
∴ Oxidation number of Al =+3