Use your textbook to determine the \"textbook value\" or \"Literature Value\" of

Question

Use your textbook to determine the "textbook value" or "Literature Value" of the Delta H values for chemical equations (1), (2) and (3) below. (2) Show equations 11) and (2) can be combined algebraically to obtain equation (3). In this experiment, you will use a Styrofoam-cup calorimeter to measure the heat released by three reactions. One of the reactions is the same as the combination of the other two reactions. Therefore, according to Hess's law, the heat of reaction of the one reaction should be equal to the sum of the heats of reaction for the other two. This concept is sometimes referred to as the additivity of heats of reaction. The primary objective of this experiment is to confirm this law. The reactions we will use in this experiment are: (1) Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s) rightarrow Na (aq) + OH^- (aq) Delta H_1? (2) Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and aqueous solution of sodium chloride. NaOH(s) + H^+(aq) + Cl^-(aq) V H_2O(l) + Na^+ (aq) + Cl^- (aq) Delta H_2 =? (3) Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride Na^+(aq) + OH^+ (aq) + Cl^- (aq) rightarrow H_2O(l) + Na^+ (aq) + Cl^-(aq) Delta H_3 =? In this experiment, you will Combine equations for two equations to obtain the equation for a third reaction. Use a calorimeter to measure the temperature change in each of three reactions. Calculate the heat of reaction, Delta H, for the three reactions. Use the results to confirm Hess's law.

Explanation / Answer

1.

NaOH(s)------->Na++OH-

H0f,NaOH=-426.73 kJ/mol

H0f,Na+=-240.1 kJ/mol

H0f,OH-=-229.99 kJ/mol

According to Hess law

H1=-240.1-229.99-(-426.73)=-43.36 kJ/mol

2.

NaOH+H++Cl------->H2O+Na++Cl-

H0f,H+=0 kJ/mol

H0f,cl-=-167.15 kJ/mol

H0f,H2O=187.78 kJ/mol

H2=-187.78-240.1-167.15-(-426.73+0-167.15)=-1.15 kJ/mol

3.

Na++OH-+H++Cl-------->H2O+Na++Cl-

H3=-187.78-229.99+0-167.15-(-426.73-240.1-167.15)=-231.14 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.