Answer : The value of
for this reaction is, 
Explanation :
The given chemical reaction is:

Now we have to calculate value of
.

![\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5Bn_%7BHCH_3CO_2%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28HCH_3CO_2%28g%29%29%7D%5D-%5Bn_%7BCH_3OH%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CH_3OH%28g%29%29%7D%2Bn_%7BCO%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CO%28g%29%29%7D%5D)
where,
= Gibbs free energy of reaction = ?
n = number of moles
= -389.8 kJ/mol
= -161.96 kJ/mol
= -137.2 kJ/mol
Now put all the given values in this expression, we get:
![\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5B1mole%5Ctimes%20%28-389.8kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%28-163.2kJ%2Fmol%29%2B1mole%5Ctimes%20%28-137.2kJ%2Fmol%29%5D)

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

where,
= standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol
R = gas constant = 8.314 J/L.atm
T = temperature = 
= equilibrium constant = ?
Now put all the given values in this expression, we get:


Thus, the value of
for this reaction is, 
Answer:
<em> 14, 508J/K</em>
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
<em>m = mass of substance (g) = 0.1184g</em>
1 mole of Mg - 24g
<em>n</em> moles - 0.1184g
<em>n = 0.0049 moles.</em>
Also, q = m × c × ΔT
<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>
<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>
<em>= 14, 508 J/K/kg</em>
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>
<em>= 1,7117.7472 J °C-1 g-1</em>
<em />
<em>∴ ΔHrxn = q/n</em>
<em>=1,7117.7472 ÷ 0.1184 </em>
<em>= 14, 508J/K</em>
Layers of rock that bend can produce a downward fold known as a syncline.
This is the type of fold that occurs with the younger layers of rocks remaining closer to the center of the structure. Synclines can be recognized by the fact that the youngest layers of rocks always remain near the folds center.
Hope this helped..