9.1 (1)

# 9.1 (1) - reaction = ∑ nΔH f(products ∑ mΔH...

This preview shows page 1. Sign up to view the full content.

Key: 1. (Try this without a calculator by using only 2 significant digits for each heat of formation) Using BLBM Table 5.3, estimate the enthalpy change for the reaction: ethene, also called ethylene (H 2 C = CH 2 ) plus hydrogen gas forms ethane (H 3 C-CH 3 ). a. 137 kJ b. - 137 kJ c. 33 kJ d. - 33 kJ e. 67 kJ f. -67 kJ g. none of these First lets refresh our understanding of what enthalpy of formation is. It is the energy required to form one mole of a certain molecule from its elements in their standard state. Enthalpies of formation can be used to predict the overall enthalpy of a reaction. This question asks for the total enthalpy change of the reaction using enthalpies of formation. The formula for these problems is ΔH
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: reaction = ∑ nΔH f (products) - ∑ mΔH f (reactants.) Note: the symbol ∑ means “the sum of”. n is the coefficient of the product. m is the coefficient of the reactants. At first this might look complicated, but it is quite simple. It just means you subtract the enthalpies of formation of the reactants from the enthalpies of formation of the products (multiplying each by its coefficient in the balanced reaction). Balanced equation: H 2 C=CH 2 + H 2 → H 3 C-CH 3 Using the balanced equation, and plugging in values from the book we get the following: 1(-84.68) – [1(52.3) + 1(0)] = -137. Since H 2 is in its standard state, its ΔH f is 0....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online