A vapour-liquid mixture of water (1) and furfural C5H4O2 (2) is maintained at 1.013 bar and 109.5 oC. Under these conditions, it is found that x1 = 0.10 and y1 = 0.81. The temperature of the mixture is changed to 100.6 oC, and some (but not all) of the vapour condenses. The vapour phase may be assumed to behave as an ideal gas. The liquid-phase activity coefficients are independent of temperature but vary with concentration according to the van Laar equations. Determine: The activity coefficients for water and furfural at 1.013 bar and 109.5 oC and the van Laar constants Vapour pressure data T= 100.6degrees p^sat1= 1.0352 bar, p^sat2=0.1193bar
T=109.5degrees p^sat1=1.4088bar p^sat2=0.1690bar

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.