Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Q6.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be ?742.7 kJ mol?1 at 298 K. Calculate enthalpy change for the reaction at 298 K. NH2CN(g) + 3\2O2(g) → N2(g) + CO2(g) + H2O(l)


Q6.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried


out in a bomb calorimeter, and ΔU was found to be ?742.7 kJ mol?1


at 298 K. Calculate enthalpy change for the reaction at 298 K.


NH2CN(g) +


 


3\2O2(g) → N2(g) + CO2(g) + H2O(l)


Grade:11

3 Answers

Aman Bansal
592 Points
9 years ago

Dear Sha,

Enthalpy change for a reaction (ΔH) is given by the expression,

ΔH = ΔU + ΔngRT

Where,

ΔU = change in internal energy

Δng = change in number of moles

For the given reaction,

Δng = ∑ng (products) – ∑ng (reactants)

= (2 – 2.5) moles

Δng = –0.5 moles

And,

ΔU = –742.7 kJ mol–1

T = 298 K

R = 8.314 × 10–3 kJ mol–1 K–1

Substituting the values in the expression of ΔH:

ΔH = (–742.7 kJ mol–1) + (–0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1)

= –742.7 – 1.2

ΔH = –743.9 kJ mol–1

Best Of Luck...!!!!

Cracking IIT just got more exciting,It’s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert

 

reetgffhghutdy
14 Points
one year ago
gdhgffhfhgfytdydgcbhj n asbmbdew fmfk cmdbcbxas xsxbgskjx sajgxajbzas sajxvj xxvjx ashxxvwxuvxjwor3hfrfn9reurek cdn cweugxkjajbxlci;exnsvcwjgs sdb kuchedjbcefkcherl e;ocjnsd sdgcwekcbelciecjsd csdncwihced, dwk gwekbsmvds sdchwihbdsdjb ad. yqocgdsm kuyoeer;jfneq vfufyeyp0e9f[e;jhfbe cdujnecvr.oyhef  n d,b d dfdfhb ab ;afoyi gajb nqygdfn dyagsv sdciuigcvdfnugvmdfjpiruoigufkrfuorequhfvr wejuytufhvrgnbv  nqbfygweneqfygvre cnjegfmhvre c eyuhdv qwfgvedgjccgggreqfhvf3owyfughcvebhfct,emnf cvfeviygverbv h
Rishi Sharma
askIITians Faculty 646 Points
11 months ago
Dear Student,
Please find below the solution to your problem.

Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH= ΔU+ ΔngRT
Where,
ΔU= change in internal energy
Δng= change in number of moles
For the given reaction,
Δng= ∑ng(products) – ∑ng(reactants)
= (2 – 2.5) moles
Δng= –0.5 moles
And,
ΔU= –742.7 kJ mol–1
T= 298 K
R = 8.314 × 10^–3kJ mol–1K–1
Substituting the values in the expression of ΔH:
ΔH= (–742.7 kJ mol–1) + (–0.5 mol) (298 K) (8.314 × 10^–3kJ mol–1K–1)
= –742.7 – 1.2
ΔH= –743.9 kJ mol–1

Thanks and Regards

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free