Are the molecules of lattice in quantum entanglement when they vibrate as a phonon? As all the molecules in the lattice are vibrating together to form a wave and the phonon is a quantum phenomenon, it makes sense to me that all the molecules inside the lattice are in the entangled state which is changing with time like propagating wave.
But it's hard to believe that so many molecules (which are in the scale of $10^{23}$) are entangled at the same time.
Are the molecules in the quantum entangled state when phonon is present in the lattice?

P.S. I'd like to add a new argument supporting the entangled molecules' view because the comments and answers claim that phonon can be explained with binding force without introducing the entanglement.
Experimenters devised a method to detect the phonon using the Raman effect.
The Raman effect is inelastic scattering. The excitation of the phonon state emits a photon and the lattice goes to the lower state during the Raman effect. The process is a quantum phenomenon. When the phonon state emits a photon, the effect has to spread through the whole lattice immediately as the phonon is collective excitation.
And I don't think the instant collapse of the wavefunction of phonon by the emission of the photon is possible only with binding force. Some sort of non-locality has to be involved and so the quantum entanglement among molecules might be the reason because entanglement is non-local.
 A: Particles participating in a wave motion are not usually entangled, and in the case of phonons in ordinary circumstances they are not.
The type of correlated motion that makes a wave is classical correlation. It is not different in principle from a mexican wave in a crowd. Each individual member of the crowd has their own individual state of motion, and this motion is correlated from one person to the next, but this is not quantum entanglement. The same goes for water molecules making a wave on the sea, and for atomic vibrations making a wave in a solid.
Quantum entanglement is happening when it is not possible, even in principle, to assign a state (e.g. a state of motion, or a spin state, or both) to each part of the system in and of itself. In technical language, the total state cannot be factorized into a product of states of subsystems. But we can easily make a wave using a product state. You just assign to each atom a state of oscillatory motion with the phase slightly advanced compared to the previous atom.
A: This is an interesting question. I would answer "no", at least not in the usual sense of quantum entanglement. A phonon is certainly a collective motion of many particles in a solid, but there are similar collective motions that are not considered entangled (e.g. plasma oscillations).
Phonons can be well described semi-classicaly: the phonon modes are vibrational modes of the solid (which can be described classically), but the energy in the modes follows a Bose-Einstein distribution (which is a result of quantum mechanics). In some sense, I think that's the definition of not entangled: something isn't entangled if it can be well described classically or semi-classically. (You might be able to make this argument more rigorous by talking about the properties of the system's density matrix.)
I realize that's kind of handwavy, but you're kind of asking a philosophical question. As far as we know, quantum mechanics applies to everything with one giant multi-particle state for the universe --- including the atoms in any solid. Arguably, everything is entangled to some degree. Nothing is truely classical. So, where do we draw the line and say that something is entangled? I'm not sure, but I think that most physicists would agree that phonons normally don't count as entangled because you can describe them fairly accurately without too much quantum mechanics.
Yes, you can move to situations where a phonon interacts with something that's definitively quantum, but such interactions are typically very brief; perhaps something that's arguably entanglement occurs --- but only for a very short period of time (short even by quantum-mechanical standards). This can often be treated as a perturbation on top of a semi-classical system.
EDIT: I should also note that you can describe phonons in quantum terms while keeping the atoms as distinguishable particles. (See J. M. Ziman's book "Electrons and Phonons" section 1.3.) The fact that we can get away with distinguishable particles also suggests a lack of entanglement between the particles.
A: 
all the molecules inside the lattice are in the entangled state

No. They are not. Many-particle entanglement is only property of Einstein-Bose condensate (BEC). For that you need to cool down gas of bosons of ultra-low density about  $1/{100~000} $ the density of normal air to ultra-low temperatures close to absolute zero of $0K$. Transition to BEC phase temperature of matter is defined as $$T_{\rm BEC}=\left(\frac{n}{\zeta(3/2)}\right)^{2/3}\frac{2\pi\hbar^2}{m k_{\rm B}}$$
Where $n$ is particle density, $\zeta$ is Riemann zeta function and $m$ is boson mass.
Also not every atom is suitable for BEC, just ones with even mass number such as ${} ^2 \text{ H} , {} ^4 \text{He} , {} ^{208}\text{Pb}$, etc. Given this, it's unlikely that solid body high-density lattice atoms in room temperature will form BEC.
