We present an exact result for the nonadiabatic transition probability and hence the defect density in the final state of a one-dimensional Kitaev model following a slow quench of the parameter J-, which estimates the anisotropy between the interactions, as J- (t) ∼- | t/τ |. Here, time t goes from - to + and τ defines the rate of change in the Hamiltonian. In other words, the spin chain initially prepared in its ground state is driven by changing J- linearly in time up to the quantum critical point, which in the model considered here occurs at t=0, reversed and then gradually decreased to its initial value at the same rate. We have thoroughly compared the reverse quenching with its counterpart forward quenching, i.e., J- ∼t/τ. Our exact calculation shows that the probability of excitations is zero for the wave vector at which the instantaneous energy gap is zero at the critical point J- =0 as opposed to the maximum value of unity in the forward quenching. It is also shown that the defect density in the final state following a reverse quenching, we propose here, is nearly half of the defects generated in the forward quenching. We argue that the defects produced when the system reaches the quantum critical point get redistributed in the wave-vector space at the final time in case of reverse quenching whereas it keeps on increasing until the final time in the forward quenching. We study the entropy density and also the time evolution of the diagonal entropy density in the case of the reverse quenching and compare it with the forward case. © 2009 The American Physical Society.