Question

need help with basic thermochemistry. PLEASE HELP

Answer 1

I got -485.6 kJ for #3

Answer 2

@aaronq

Answer 3

that's right, i posted that calculation in the previous thread - you must've missed it

Answer 4

yea u did, I just didnt go back

Answer 5

I'm doing 4 and I'm lost @aaronq there is more products in the equations than I need for the target equation

Answer 6

Calculate ∆H for the reaction CH4 (g) + NH3 (g) \(\rightarrow\) HCN (g) + 3 H2 (g), from the reactions.

1. N2 (g) + 3 H2 (g) \(\rightarrow\) 2 NH3 (g) ∆H = –91.8 kJ
2. C (s, graphite) + 2 H2 (g) \(\rightarrow\) CH4 (g) ∆H = –74.9 kJ/mol
3. H2 (g) + 2 C (s, graphite) + N2 (g) \(\rightarrow\) 2 HCN (g) ∆H = +270.3 kJ

to 1.: divide by 2 and reverse
to 2.: reverse
to 3.: divide by 2



1. NH3 (g) \(\rightarrow\) 1/2 N2 (g) + 3/2 H2 (g)
2. CH4 (g) \(\rightarrow\) C (s, graphite) + 2 H2 (g)
3. 1/2 H2 (g) + C (s, graphite) + 1/2 N2 (g) \(\rightarrow\) HCN (g)

1.∆H =(1/2)*(+270.3) kJ
2.∆H = 74.9 kJ/mol
3. ∆H =(1/2)*91.8 kJ

Answer 7

1. NH3 (g) → 1/2 N2 (g) + \(\cancel{1/2 H2 (g)}\)+ 2/2 H2 (g)
2. CH4 (g) → C (s, graphite) + 2 H2 (g)
3. \(\cancel{1/2 H2 (g)}\) + C (s, graphite) + 1/2 N2 (g) → HCN (g)

etc

Answer 8

hmm

Answer 9

doubts?

Answer 10

yea trying to figure it out on paper

Answer 11

okay cool

Answer 12

for 2, why isnt the C(s) cancelled out? @aaronq

Answer 13

it should be, i just didn't do all of them to keep things focused.

the 1/2 N2 and the C should be cancelled

Answer 14

ok

Answer 15

@sweetburger I have a quick question

Answer 16

thers like 4 on that pdf which one @Anikate

Answer 17

5

Answer 18

Im wondering if I have to flip any equations in that question

Answer 19

You have to reverse the last one for sure.

Answer 20

and I think you have to multiply the last one by 2 but im not sure yet

Answer 21

I think you mulitply the 3rd by 3 and the 2nd by 6

Answer 22

1. nothing
2. multiply by 6
3. multiply by 3
4. reverse and mutlipyl by2
let me check this to see if it works

Answer 23

sorry my computer shut off. im back

Answer 24

ya so the steps i listed should be correct that will allow you to find the enthalpy change

Answer 25

if you have to reverse 4 dont u also have to reverse 2?

Answer 26

no I dont think so?

Answer 27

why would 2 have to be reversed?

Answer 28

because HCl is in the equation

Answer 29

just like AlCL3 is in the equation

Answer 30

@sweetburger

Answer 31

Ok so if you reverse 4 your changing which side the solid and gaseous version of AlCl3 is. This doesnt affect what happens to #2 reaction. Literally apply that changes the i told you to do and then cancel out and see what is left.

Answer 32

oh ok do I only reverse an equation if one of its reactants is suppsoed to be a product?

Answer 33

is that true @sweetburger

Answer 34

@Anikate Im almost certain

Answer 35

sweet i have to take an online chem quiz on thermo basics, can you pleas help me?

Answer 36

its 25 questions

Answer 37

timed 40 mins

Answer 38

you there? @sweetburger

Answer 39

are you busy right now?

Answer 40

bro Im busy right now I dont want to say i can be here and help and have you rely on me if I have to go

Answer 41

oh ok, so can you help? imma need instand replies just double my work and answer questions

Answer 42

@mathstudent55