Wait what
What
Huh
Kinda confused
Answer: 5.85kJ/Kmol.
Explanation:
The balanced equilibrium reaction is
The expression for equilibrium reaction will be,
![K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5Bp_%7BD%7D%5D%5Ctimes%20%5Bp_%7BC%7D%5D%7D%5E4%7B%5Bp_%7BB%7D%5D%5E2%5Ctimes%20%5Bp_%7BA%7D%5D%7D)
Now put all the given values in this expression, we get the concentration of methane.
Relation of standard change in Gibbs free energy and equilibrium constant is given by:
where,
R = universal gas constant = 8.314 J/K/mole
T = temperature = 
= equilibrium constant = 10.6
Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.
Answer:fH = - 3,255.7 kJ/mol
Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
Consider the following equation:0.0136 g + 2.70 × 10-4 g - 4.21 × 10-3 g = ? How many digits to the right of the decimal point should be used to report the result? =2, Choose the correct answer=9.66 × 10^-3 g, Which metric unit would be the best choice to report the result?=mg
