Entropy change in an irreversible process between 2 equilibrium state Calculating entropy change in an irreversible process between 2 states requires computing the change in entropy for any reversible process between the 2 same states, but why?
If someone could provide a simple explanation for the above concept for me to build a frame work upon, that would be really good. 
 A: Entropy is a state function which means that the value of entropy change between 2 states depends only on the initial and final states and not the path/process taken from the initial state to reach the final state. 
I'm not sure about this part but I think entropy calculations are more easier for reversible processes than irreversible ones.
A: Entropy of a system can increase either by entropy transfer or due to entropy generation. If we live in a world where every process is reversible then entropy will never be generated only transfered from one system to other. Entropy generation is due to irreversibilities in a process.Lets consider water taken inside an adiabatic container we do some work on the system by a stirer..when we stir water the entropy of the system increases. But there is no entropy transfer associated with work .Then how does the entropy of the system increase ? Due to internal irreversibility called viscocity. now here entropy hasn't been transfered from one system to other instead entropy has been generated. Now how will you calculate the entropy generated. We know that entropy is a state function. It doesn't matter if the entropy increase is due to generation or transfer. So all we need is the initial and final state of the system and a reversible path..Here we will take reversible heat transfer as our path..So instead of taking the work we will imagine a same amount heat transfered into the system in a reversible manner . Now we know the equation for entropy transfer associated with reversible heat transfer
. So if T1 and T2 is the initial and final temprature. Entropy change
                            S=integral of mcdT/T  taking limits T2 and T1 
            here m is the mass and c the specific heat of water. So you see why we needed the reversible path. we know how to calculate entropy transfer but not the entropy generated .Both the paths are equiliant . In both the paths entropy change and energy change are the same. The only difference is entropy was generated in first case but it was transfered in second case      
A: As @Binary Geek states, the entropy change is not dependent on the path/process but only on the initial and final states. The reason is that entropy is a state variable (also called a state function). 
For reversible processes, the entropy change can be calculated through integration of the second law of thermodynamics for reversible processes:
$$dS = \frac{\delta Q}{T}$$
Note that the equality sign applies only because of the assumption that the process is reversible. Otherwise, for irreversible processes, we have the greater-than sign, and $dS$ can't be calculated.
Summary
It doesn't matter what kind of process you use to calculate entropy change, but choosing a (sequence of) reversible process(es) is the easiest, if not the only option.
