Question

Another question

Answer 1

Question is attached

Answer 2

@sweetburger @Rushwr

Answer 3

is it C ?

Answer 4

\[2CH _{4} + 3O _{2} --> 2CO + 4H _{2}O\]

Answer 5

Like i though that for this reaction we have entropy goes up because we have

on the reactant side

2+3 moles = 5 moles

and on the product side we have

2+4 = 6 moles.

So we have more moles on the product side so i thought more moles = more disorder.

Answer 6

\[\Delta G = \Delta H - T \Delta S\]

Answer 7

wait... entropy >0 and if it's spontaneous G<0 so h <0

Answer 8

aaaah yes

Answer 9

okai since the product side had more moles than the reactants we took \[\Delta S\] as pistive right? Okai so anyhow the T\[\Delta S\] we consider the whole part as negative. But to find the enthalpy change if its +ve or -ve we need to know if the reaction is spontaneous or not right? But they haven't given that here.

Answer 10

@Rushwr I agree, I had trouble with that, actually I just assumed that the reaction was spontaneous.

b/c then I knew okay delta S was positive
so then I thought that delta H had to be negative

the book doesnt really give an explanation as to how you can figure out that it's exothermic, like it just says oh " this type of reaction is exothermic"

Answer 11

yeah so this reaction is exothermic

Answer 12

bond formation is always exothermic while bond dissociation is always endothermic. hence this reaction is exothermic. In exothermic reactions delta H is positive. Cuz it produces heat while in endothermic reactions delta H is negative cuz they absorb heat ! So that is why I choose that one