e are all familiar with the general principles of operation of an internal combustion reaction: the combustion of fuel drives out the piston. It is possible to imagine engines that use reactions other than combustions, and we need to assess the work they can do. A chemical reaction takes place in a container of cross-sectional area 100 cm2 ; the container has a piston at one end. As a result of the reaction, the piston is pushed out through 10.0 cm against a constant external pressure of 100 kPa. Calculate the work done by the system.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

The mole ratio of hydrogen to ammonia is 3: 2.

Seems cool i guess ttyl

Reaction involved in present system is :   c12h22o11 + h2o → 2 c6h12o6the reaction system is obeying following rate law:   rate = k[c12h22o11]thus, it is a first order reaction.now, for 1st order reactionrate constant (k) =  given: initial conc. = 3.0 g, final conc. = 2.7 g, t = 2.5 h∴rate constant (k) =  ∴ k = 0.04215 h-1now, for 1st order reaction, half life = 0.693/k = 0.693/0.04215 = 16.4 hthus correct answer is option b