The production of methyl bromide is an irreversible liquid-phase reaction that follows an elementary rate law. the reaction cnbr (a) + ch3nh2 (b) à ch3br (c) + ncnh2 (d) is carried out isothermally in a semi-batch reactor. a solution of methylamine (b) in extra dry ethanol at a concentration of 0.025 mol/l is to be fed at a rate of 0.05 l/s to an solution of bromine cyanide (a) in extra dry ethanol contained in a glass-lined reactor. the initial volume of fluid in the reactor is to be 5 l with a bromine cyanide concentration of 0.05 mol/l. the specific reaction rate constant is

the answer to your question is c

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.