Plz explain 4th question... te +0, ()= co (g) at 298 k and 1 atm, ah =- 264lb if the molar

please ask if any doubt in this😊

Vfinal we calculate here shweta...as molar volume is 24.46 and change in number of gaseous moles =1-1/2=1/2

V initial of graphite is given to you

P is 1 atm and then calculate all these values

sir

graphite molar volume is given

yes shweta this is V initial reactant side

after completion at 298K and 1 atm..molar volume is 24.47(general)

we calculate Vfinal

change in gaseous moles * 24.47

got it?

22 .4 l sir

At 25°C and 1 atm 1 mole of gas occupies 24.47 molar volume

o

At 0°C and 1 atm it occupy 22.4L

k

sir

😊🙂👍

sir

am getting -3.458 kj sir ..

this is next question

okk let me check shweta

ok sir

shweta 2 ques is pending after answering this i will solve this and send you here

okay sir🙂

just tell me one thing

can I solve that 4 th question like this

shweta formula is right but ans is also not coming....here Volume is given you have to use formula in which volume is there

sir but m not getting why u have multiply Ng factor

there

for final process in finding V final

you took 25c in place of 298 k.. put it the eq. answer is coming to be 26714

sorry its dH = dE + dnRT

yes ..

thanks

sir

mai mail pe bhej deta hu

sir

sir

got it??

what are the values of vfinal and v initial

app is not working well

DH = DE + PDV, in order to calculate DE we need to calculate PDV. The variation of volume is due both to the disappearance of solid graphite and to the variation of number of moles (1 - 1/2 = 0.5). Remembering that a mole of a gas at standard condition (0 oC and 1 atm) occupies 24.4 liters: DV = (0.5) (24.4) = 12.2 liter Thus the variation of volume due to the variation of number of moles is 12.2 and, by neglecting the variation of volume due to the disappearance of solid graphite (0.0053 liter), PDV = (1 atm)(12.2 liter) = 12.2 liter-atm Since 1 liter-atm is 24.4 cal,  PDV = (12.2) (24.4) = 297.68 cal DH = DE + PDV DE = -26416 - 297.68 = -26713 cal Alternatively and more simple, remembering that PDV = DnRT and that Dn = 0.5 PDV = 0.5 RT = (O.5)(1.987) (298) = 296 DE = -26416 - 296 = 26712 cal In this last case, care must be taken in the choice of R (here cal/mole deg).

here D = delta