(x) Calculate the work done during the reaction represented bythe foll

1.	(x) Calculate the work done during the reaction represented bythe following thermochemical equation at 300 K:CH4 +202 → CO2 + 2H2O(g) (g) (g) (1)(R=8-314 JK-'mol-!)(a) +4.988 kJ(b). - 4.988 kJ(c) - 49.88 kJ(d) + 49.88 kJ
Answer» Given: Thermochemical equation CH4 (g) +202 (g) → CO2 (g) + 2H2O (l) To FIND: Work done during the rection. Solution: As we have given the thermochemical equation, so we can identify that the equation is already balanced. So no need to balance again. Now, we know the formula of Δn which is: Δn = no. of gaseous product - number of gaseous REACTANT = 1 - (1+2) = -2 Now putting all the values in the formula of work done, we get: W = ΔnRT = -2 ( 8.314) (300) = 4988.4 J W = 4.9884 KJ Answer: The work done during the REACTION is 4.9884 KJ

(x) Calculate the work done during the reaction represented bythe following thermochemical equation at 300 K:CH4 +202 → CO2 + 2H2O(g) (g) (g) (1)(R=8-314 JK-'mol-!)(a) +4.988 kJ(b). - 4.988 kJ(c) - 49.88 kJ(d) + 49.88 kJ

Answer»

Given: Thermochemical equation CH4 (g) +202 (g) → CO2 (g) + 2H2O (l)

To FIND: Work done during the rection.

Solution:

As we have given the thermochemical equation, so we can identify that the equation is already balanced. So no need to balance again.
Now, we know the formula of Δn which is:

Δn = no. of gaseous product - number of gaseous REACTANT

= 1 - (1+2)

= -2

W = ΔnRT

= -2 ( 8.314) (300)

= 4988.4 J

W = 4.9884 KJ

Answer:

The work done during the REACTION is 4.9884 KJ