Midterm Review #12b

Moderators: Chem_Mod, Chem_Admin

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

Midterm Review #12b

Postby Brian Tangsombatvisit 1C » Sun Feb 09, 2020 11:08 pm

Suppose a researcher finds that delta Hrxn = -2756 kJ for the reaction at 200. C. Assuming all heat capacities are constant, calculate delta Hrxn at the temperature of the human body, 37 C. Hint: since enthalpy is a state function, the process can be divided into three steps.

Can somebody help me on how to start this problem? What are the three steps we can divide this process into?

Kaitlyn Ang 1J
Posts: 116
Joined: Fri Aug 09, 2019 12:17 am
Been upvoted: 1 time

Re: Midterm Review #12b

Postby Kaitlyn Ang 1J » Sun Feb 09, 2020 11:20 pm

For this problem, you have 3 mini reactions. The first is the rise in temperature of the reactants. The second is the change from reactants to products at the higher temperature. The third is the decrease in temperature of the products.
In other words, the three steps are

1) the enthalpy change of the reactants from 37C to 200C
2) the enthalpy change of the reactants to products at 200C (this is given to you by the fact that the reaction is -2756 kJ)
3) the enthalpy change of the products from 200C back down to 37C

You would calculate steps 1 and 3 with the equation of deltaH = nCdeltaT and use the corresponding C to the specific reactant/product.

Brian Tangsombatvisit 1C
Posts: 119
Joined: Sat Aug 17, 2019 12:15 am

Re: Midterm Review #12b

Postby Brian Tangsombatvisit 1C » Sun Feb 09, 2020 11:59 pm

would the enthalpies for the reactants and products in steps 1 and 3 be both positive? or would reactants be positive and products be negative?

Daria Azizad 1K
Posts: 116
Joined: Thu Jul 25, 2019 12:15 am

Re: Midterm Review #12b

Postby Daria Azizad 1K » Mon Feb 10, 2020 12:31 am

Brian Tangsombatvisit 1C wrote:would the enthalpies for the reactants and products in steps 1 and 3 be both positive? or would reactants be positive and products be negative?

if the calculation for q in step 1 is positive because final temp is higher than initial and in step 3 it is negative because final temp is lower than initial


Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

Who is online

Users browsing this forum: No registered users and 1 guest