Home
Class 12
CHEMISTRY
Calculate standard free energy change fr...

Calculate standard free energy change from the free energy of formation data for the following reaction, `C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to 6CO_(2)(g)+3H_(2)O(g)`.
`Delta_(f)G^(theta)[C_(6)H_(6)(l)]=172.8` kJ/mol
`Delta_(f)G^(theta)[CO_(2)(g)]=-394.4` kJ/mol

`Delta_(r )G^(theta)[H_(2)O(l)]=-228.6` kJ/mol

Text Solution

AI Generated Solution

To calculate the standard free energy change (ΔG°) for the given reaction: \[ C_{6}H_{6}(l) + \frac{15}{2}O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(g) \] we will use the standard free energy of formation (ΔfG°) values provided for each species involved in the reaction. ### Step-by-step solution: ...
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL THERMODYNAMICS

    VMC MODULES ENGLISH|Exercise SOLVED EXAMPLES|19 Videos
  • CHEMICAL THERMODYNAMICS

    VMC MODULES ENGLISH|Exercise PRACTICE EXERCISE|22 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

Calculate the standard Gibbs free energy change from the free energies of formation data for the following reaction: C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(g) Given that Delta_(f)G^(Theta) =[C_(6)H_(6)(l)] = 172.8 kJ mol^(-1) Delta_(f)G^(Theta)[CO_(2)(g)] =- 394.4 kJ mol^(-1) Delta_(f)G^(Theta) [H_(2)O(g)] =- 228.6 kJ mol^(-1)

Consider the following reaction. C_(6)H_(6)(l)+(15)/(2)O_(2)(g)rarr6CO_(2)(g)+3H_(2)O(g) signs of DeltaH, DeltaS and DeltaG for the above reaction will be

Consider the reaction: 4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l) DeltaG^(Theta) =- 1010.5 kJ Calculate Delta_(f)G^(Theta) [NO(g)] if Delta_(f)G^(Theta) (NH_(3)) = -16.6 kJ mol^(-1) and Delta_(f)G^(Theta) [H_(2)O(l)] =- 237.2 kJ mol^(-1) .

Determine the enthalpy of formation of B_(2)H_(6) (g) in kJ/mol of the following reaction : B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(3)(s)+3H_(2)O(g) , Given : Delta_(r )H^(@)=-1941 " kJ"//"mol", " "DeltaH_(f)^(@)(B_(2)O_(3),s)=-1273" kJ"//"mol," DeltaH_(f)^(@)(H_(2)O,g)=-241.8 " kJ"//"mol"

Calculate the standard internal energy change for the following reactions at 25^(@)C : 2H_(2)O_(2)(l) rarr 2H_(2)O(l) +O_(2)(g) Delta_(f)H^(Theta) at 25^(@)C for H_(2)O_(2)(l) =- 188 kJ mol^(-1) H_(2)O(l) =- 286 kJ mol^(-1)

At 27^(@)C for reaction, C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2)O(l) proceeds spontaneously because the magnitude of

For the reaction : C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(g) if Delta U^(@)= -1373 kJ mol^(-1) at 298 K . Calculate Delta H^(@)

In the reaction C_(3)H_(6)(g)+nO_(2)(g)toCO_(2)(g)+H_(2)O(l) . The ratio of the coefficients of CO_(2) and H_(2)O is

The fat, C_(57)H_(104)O_(6)(s) , is metabolized via the following reaction. Given the enthalpies of formation, calculate the energy (kJ) liberated when 1.0 g of this fat reacts. C_(57)H_(104)O_(6)(s)+80 O_(2)(g)rarr57CO_(2)(g)+52 H_(2)O(l) Delta_(f)H^(@)(C_(57)H_(104)O_(6),s)=-70870" kJ"//"mol, "Delta_(f)H^(@)(H_(2)O,l)=-285.8 " kJ"//"mol" , Delta_(f)H^(@)(CO_(2),g)=-393.5 " kJ"//"mol"

Calculated the equilibrium constant for the following reaction at 298K : 2H_(2)O(l) rarr 2H_(2)(g) +O_(2)(g) Delta_(f)G^(Theta) (H_(2)O) =- 237.2 kJ mol^(-1),R = 8.314 J mol^(-1) K^(-1)