Answer:
O-H bond
Explanation:
Let us work out the electronegativity difference between the elements in each bond in order to decide which of them is most polar.
For the C-O bond
2.55 - 2.2 =0.35
For the F-F bond
3.98 - 3.98 = 0
For the O-H bond
3.44 - 2.2 = 1.24
For the N-H bond
3.04 - 2.2 = 0.84
The O-H bond has the highest electronegativity difference, hence it is he most polar bond.
Answer:
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
Explanation:
2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)
Let's apply the thermodynamic formula to calculate the ΔG
ΔG = ΔG° + R .T . lnQ
We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q
How can we know Q? By the partial pressures (Qp)
P NO = 0.450atm
PO₂ = 0.1 atm
PNO₂ = 0.650 atm
Qp = [NO₂]² / [NO]² . [O₂]
Qp = 0.650² / 0.450² . 0.1 = 20.86
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
Answer:
C) In[reactant] vs. time
Explanation:
For a first order reaction the integrated rate law equation is:
where A(0) = initial concentration of the reactant
A = concentration after time 't'
k = rate constant
Taking ln on both sides gives:
![ln[A] = ln[A]_{0}-kt](https://tex.z-dn.net/?f=ln%5BA%5D%20%3D%20ln%5BA%5D_%7B0%7D-kt)
Therefore a plot of ln[A] vs t should give a straight line with a slope = -k
Hence, ln[reactant] vs time should be plotted for a first order reaction.
The last one would be false
Answer:
are those the answer choice
