Question

Consider the reaction of glucose with oxygen: C6H12O6(s) + 6O2(g) Right arrow. 6CO2(g) + 6H2O(l)

The enthalpy of formation (Delta.Hf) for C6H12O6(s) is –1,273.02 kJ/mol, Delta.Hf for CO2(g) is –393.5 kJ/mol, and
Delta.Hf for H2O(l) is –285.83 kJ/mol. What is Delta.Hf for O2(g)?
✔ exactly 0 kJ/mol

Which equation should be used to calculate Delta.Hrxn for this reaction?

(-393.5kJ/mol + (-285.83 kJ/mol))-(-1,273.02 kJ/mol)

-1,273.02 kJ/mol - ((6 mol)(-393.5 kJ/mol) + (6 mol)(-285.83 kJ/mol))
((6 mol)(-393.5 kJ/mol) + (6 mol)(-285.83 kJ/mol)) - (1 mol)(-1,273.02 kJ/mol)

What is Delta.Hrxn for this reaction?

Answer 1

Answer:

Subtract the sum of the heats of formation of the reactants from that of the products to determine delta H: delta H = –110.53 kJ/mol – (–285.83 kJ/mol) = 175.3 kJ.

Explanation:

Step 1: Set Up the Equation. Arrange your given ΔHf and ΔH values according to the following equation: ΔH = ΔHf (products) - ΔHf (reactants). ...

Step 2: Solve the Equation. Solve your equation for ΔHf. ...

Step 3: Validate the Sign. Adjust your ΔHf value's sign depending on whether it is for a product or a reactant.

Answer 2

The heat of reaction is given by the following;((6 mol)(-393.5 kJ/mol) + (6 mol)(-285.83 kJ/mol)) - (1 mol)(-1,273.02 kJ/mol)

The heat of reaction presence of oxygen. Given that we have the information provided in the question, it then follows that we can calculate the heat of the reaction using the equation;  ((6 mol)(-393.5 kJ/mol) + (6 mol)(-285.83 kJ/mol)) - (1 mol)(-1,273.02 kJ/mol)

Hence the enthalpy of reaction can be obtained from the heats of formation of the reactants according to the balanced reaction equation.

Learn more about  enthalpy of reaction: https://brainly.com/question/1657608