Afirst-order reaction has a half-life of 22.9 s . how long does it take for the concentration of the reactant in the reaction to fall to one-sixteenth of its initial value

The half-life of a first order reaction is in how much τtime, the concentration of the reactant halves. After one half-life, the concentration of the reactant becomes half its initial value. After another half-life, itsconcentration becomes (M: initial concentration):
In general, after n half-lifes, the concentration of the reactant becomes: 
Since 1/16= , we have that 4 half-lifes need to pass so that the concentration of the reactant becomes 1/16 of the initial one (=1/16*M). Hence, 4*22.9 sec=91.6 seconds is the required time.

Any photo? or something

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.