(a) A 1.00 mole sample of an ideal gas in a vessel with a movable piston initially occupies a volume of 4.00 L at an external pressure of 6.00 atm. If the external pressure is suddenly lowered to 2.00 atm and the gas is allowed to expand isothermally, calculate the following quantities for the system: ?E, w, q, ?H, ?S, and ?G (Note: 1 atm L = 101.3 J) (b) The standard enthalpy of ice melting is ?H?= 6.03×103 J/mol at its normal melting point of 0°C. What is the standard entropy of ice melting (?S°)? Show by the second law (i. e., comparing the entropy changes of the system and the surrounding) that this process will occur spontaneously at a temperature of 10 °C .

is it still possible that part a can be solved?

Explanation:

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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.