UFRJ | Leaf litter decomposition experiment in two phytophysiognomies of a sandbank in northern Rio de Janeiro State, Brazil

This technique was chosen because it is widely used and has been described in other studies, which will be discussed later.A total of 236 litterbags were made, including 15 sets consisting of 12 litterbags with 5-mm mesh and 14 sets consisting of 4 litterbags with 0.02-mm mesh. Thus, 5 thickets in the FAAC dominated by C. hilariana and 5 not dominated by C. hilariana received a set of 12 litterbags with a 5-mm mesh. The Periodically Flooded Forest received 5 sets consisting of 12 litterbags with a 5-mm mesh and 4 sets consisting of 4 litterbags with a 0.02-mm mesh.

The sets were assembled so that the litterbags for each type of vegetation formation contained their respective litter.Cada conjunto de malha 5mm foram despositados 12 litterbags, totalizando180 litterbags, e em cada conjunto de malha 0,02mm disposta na FAAC foi composto por 4 litterbags, totalizando 56 litterbags.The litterbag assemblies were installed in the field in November 2012 and remained exposed to the elements and macroinvertebrates for 24 months.

A 5-mm mesh litterbag was retrieved at regular intervals at 59, 119, 169, 226, 282, 337, 407, 450, 541, 594, 649, and 710 days after installation, while the 0.02-mm mesh litterbags were removed at 131, 253, 380, and 519 days after installation.The remaining leaf litter from each litterbag was processed to remove debris (soil, fine roots, soil fauna), dried at approximately 50°C for 72 hours, and weighed on an analytical balance.

The weight of the remaining material in each litterbag, from the time of installation (t=0) until collection, was determined based on the calculation of the decomposition constant (k), according to Oslon’s 1963 equation.Using the value of k, the half-life of the leaves was calculated for the FAAC, MSC, and MCC areas, as well as for the MPI. To determine the time required for 50% of the dry weight to be reduced, the following formula was used: -ln (0.5) /k = 0.693/k, so t₀.₅ = 0.693/k.

To determine the time required for a 95% loss of material, the calculation used 3/k (Olson, 1963).