The standard enthalpy of formation of a compound is defined as the change in enthalpy that accompanies the formation of one mole of the compound from its elements in their standard states. The standard state of an element is its most stable state at 1 atm and 25°C.

Looking at the options:

A. 6C (s) + 6H 2(g) + 3O2(g) → C 6H12O6(s)
B. 6CO2(g) + 6H2O (l) → C6H12O6(s) + 6O 2(g)
C. 6C(g) + 6H 2(g) + 3O2(g) → C6H12O 6(s)
D. 6C(g) + 6H 2(g) + 3O2(g) → C 6H 12O 6(aq)

The correct answer is A. 6C (s) + 6H 2(g) + 3O2(g) → C 6H12O6(s). This is because the standard enthalpy of formation involves the formation of one mole of the compound from its elements in their standard states. In this case, carbon (C) is in its standard state as a solid (s), hydrogen (H2) is in its standard state as a gas (g), and oxygen (O2) is in its standard state as a gas (g). These react to form one mole of glucose (C6H12O6) in its standard state as a solid (s).

Question

The standard enthalpy of formation of a compound is defined as the change in enthalpy that accompanies the formation of one mole of the compound from its elements in their standard states. The standard state of an element is its most stable state at 1 atm and 25°C.

Looking at the options:

A. 6C (s) + 6H 2(g) + 3O2(g) → C 6H12O6(s)
B. 6CO2(g) + 6H2O (l) → C6H12O6(s) + 6O 2(g)
C. 6C(g) + 6H 2(g) + 3O2(g) → C6H12O 6(s)
D. 6C(g) + 6H 2(g) + 3O2(g) → C 6H 12O 6(aq)

The correct answer is A. 6C (s) + 6H 2(g) + 3O2(g) → C 6H12O6(s). This is because the standard enthalpy of formation involves the formation of one mole of the compound from its elements in their standard states. In this case, carbon (C) is in its standard state as a solid (s), hydrogen (H2) is in its standard state as a gas (g), and oxygen (O2) is in its standard state as a gas (g). These react to form one mole of glucose (C6H12O6) in its standard state as a solid (s).

Knowee AI · Accepted Answer

The standard enthalpy of formation of a compound is defined as the change in enthalpy that accompanies the formation of one mole of the compound from its elements in their standard states. The standard state of an element is its most stable state at 1 atm and 25°C.

Looking at the options:

A. 6C (s) + 6H 2(g) + 3O2(g) → C 6H12O6(s)
B. 6CO2(g) + 6H2O (l) → C6H12O6(s) + 6O 2(g)
C. 6C(g) + 6H 2(g) + 3O2(g) → C6H12O 6(s)
D. 6C(g) + 6H 2(g) + 3O2(g) → C 6H 12O 6(aq)

The correct answer is A. 6C (s) + 6H 2(g) + 3O2(g) → C 6H12O6(s). This is because the standard enthalpy of formation involves the formation of one mole of the compound from its elements in their standard states. In this case, carbon (C) is in its standard state as a solid (s), hydrogen (H2) is in its standard state as a gas (g), and oxygen (O2) is in its standard state as a gas (g). These react to form one mole of glucose (C6H12O6) in its standard state as a solid (s).

Which of the following set of equations could be used to represent the standard enthalpy of formation of glucose?A. 6C (s) + 6H 2(g) + 3O2(g) → C 6H12O6(s)B. 6CO2(g) + 6H2O (l) → C6H12O6(s) + 6O 2(g)C. 6C(g) + 6H 2(g) + 3O2(g) → C6H12O 6(s)D. 6C(g) + 6H 2(g) + 3O2(g) → C 6H 12O 6(aq)

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