Home
Class 12
CHEMISTRY
Given : C("diamond")+O(2)rarrCO(2),De...

Given :
`C("diamond")+O_(2)rarrCO_(2),DeltaH=-395 kJ`
`C("graphite")+O_(2)rarrCO_(2),DeltaH=-393 kJ`
The enthalpy of formation of diamond from graphite is

A

`-1.5 kJ`

B

`+1.5 kJ`

C

`+3.0 kJ`

D

`-3.0 kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL THERMODYNAMICS

    VMC MODULES ENGLISH|Exercise IN - CHAPTER EXERCISE - I|10 Videos
  • CHEMICAL THERMODYNAMICS

    VMC MODULES ENGLISH|Exercise IN - CHAPTER EXERCISE - J|10 Videos
  • CHEMICAL THERMODYNAMICS

    VMC MODULES ENGLISH|Exercise IN - CHAPTER EXERCISE - G|10 Videos
  • CHEMICAL KINETICS

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|52 Videos
  • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

    VMC MODULES ENGLISH|Exercise JEE ADVANCE (ARCHIVE)|30 Videos

Similar Questions

Explore conceptually related problems

Given that C+O_(2)rarrCO_(2),DeltaH^(@)=-xKJ and 2CO+O_(2)rarr2CO_(2),DeltaH^(@)=-yKJ The enthalpy of formation of carbon monoxide will be

For the reaction, C+O_(2) rarr CO_(2),DeltaH=-393 J 2Zn + O_(2) rarr 2ZnO, DeltaH=-412 J

Given that: i. C("graphite") +O_(2)(g) rarr CO_(2)(g), DeltaH =- 393.7 kJ ii. C("diamond") rarr C("graphite"), DeltaH =- 2.1 kJ a. Calculate DeltaH for buring of diamond of CO_(2) . b. Calculate the quantity of graphite that must be burnt to evolve 5000 kJ of heta.

Based on the following reaction C(graphite) + O_2(g)rarrCO_2(g),DeltaH=-394KJ//mol...(i) 2CO(g)+O_2(g)rarr2CO_2(g),DeltaH=-569KJ//mol...(ii) The heat of formation of CO will be

Determine the heat of transformaiton of C_(("dimamond"))rarrC_(("graphite")) form the following data: i. C_(("diamond"))+O_(2)(g) rarr CO_(2)(g), DeltaH^(Theta) =- 94.5 kcal ii. C_(("graphite")) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 94.0 kcal

Determine the heat of transformaiton of C_(("dimamond"))rarrC_(("graphite")) form the following data: i. C_(("diamond"))+O_(2)(g) rarr CO_(2)(g), DeltaH^(Theta) =- 94.5 kcal ii. C_(("graphite")) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 94.0 kcal

Two reactions are given below : C_("(graphite)")+O_(2(g))rarrCO_(2(g)),DeltaH=-393.7kJ C_("(diamond)")rarrC_("(graphite)"),DeltaH=-2.1kJ What quantity of diamond will give 800 kJ of heat on burning ?

On the basis of the following thermochemical data : (Delta_(f)G^(@)H_((aq.))^(+)=0) H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ The value of enthalpy of formation of OH^(-) ion at 25^(@)C is :

Form the thermochemical reactions. C(graphite) +1//2 O_(2)(g) to CO(g), triangleH=-110.5kJ CO(g)+1//2O_(2)(g) to CO_(2)(g), triangleH=-283.2kJ The enthalpy ofreaction of C(graphite) +O_(2)(g) to CO_(2)" is "

If S+O_(2)toSO_(2),DeltaH=-298.2 " kJ" " mole"^(-1) SO_(2)+(1)/(2)O_(2)toSO_(3)DeltaH=-98.7 " kJ" " mole"^(-1) SO_(3)+H_(2)OtoH_(2)SO_(4),DeltaH=-130.2 " kJ" " mole"^(-1) H_(2)+(1)/(2)O_(2)toH_(2)O,DeltaH=-287.3 " kJ" " mole"^(-1) the enthlapy of formation of H_(2)SO_(4) at 298 K will be