Chemistry 1302A/B Lecture Notes - Lecture 6: Calorimetry, Mols

31 views3 pages

blushporpoise372

10 Feb 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

How much heat energy is required to take 22. 0g of ice at -15. 0 c to water vapour art. Step 1: heat ice from -15 c 0 c q = m x c x t. =22. 0g x 2. 06jg-1c-1 x 15 c = 679. 8 j. Step 2: melt ice at 0 c q = mols x lh. Step 3: heat water from 0 c 100 c q= m x c x t. Step 4: vaporize at 100 c q = mols x lh. Step 5: heat vapour from 100 c c q = m x c x t. When heat is evolved, system has lost energy and enthalpy has decreased: h < 0 reaction is exothermic. When heat is absorbed, system has gained energy and h > 0 reaction is endothermic. Calorimetry is a container used to measure heat transfer in a chemical process. It operates at a constant pressure so that: qrxn = h rxn.

Related Questions

Show transcribed image text

taupepug111

Compound Î Hrxn (kJ/mol) NH4NO3 + 25.7 KCl + 17.2 NaOH -44.5 KOH -57.6

You are given 6.7 grams of an unknown salt. You dissolve it in 54.1 mL of water. As the salt dissolves, the temperature of the solution changes from an initial temperature of 26.9Â°C to a final temperature of 53.9Â°C. Based on the chart of ÎH values and the temperature change, you can determine which salt you have. What is the temperature change? Î T = o C Tries 0/99

This is the ÎT component of the expression: qsurr = C x mass x ÎT where qsurr is the heat lost or gained by the contents of the calorimeter due to the chemical reaction. What is the mass that is changing temperature in this experiment? mass =_______ g

Assume that the heat capacity, C, of the solution is the same as that of water, C = 4.184 J/g oC. Solve for qsurr. qsurr = J Tries 0/99 What is the value of qsys? qsys = ____J

The change in enthalpy, ÎH, of this reaction is equal to qsys, since it was carried out under conditions of constant external pressure. Based on the sign of ÎH (or qsys) the dissolution of the unknown salt is . ________

When 6.7 g of salt are dissolved in 54.1 mL of water, 6880.4 J of heat are given off by the system. How much heat would be given off if only 1.00 g of salt had been dissolved in the water? qsys =_______ J/g

In the table, the ÎH values are given in kJ/mol. Convert your last answer to kJ to give kJ/g. Finally, multiply the kJ/g by the molar masses (g/mol) of possible salts, to get kJ/mol. The reaction of your unknown salt with water was exothermic. So, you will only need to look at salts with negative ÎH values. formula of unknown = ________

Chemistry 1302A/B Lecture Notes - Lecture 6: Calorimetry, Mols

Document Summary

Get access

Related Documents

Chemistry 1302A/B Study Guide - Midterm Guide: Ideal Gas Law, Reaction Quotient, Molar Volume

Chemistry 1302A/B Chapter Notes - Chapter 2.2: Pi Bond, Bond Order, Sigma Bond

CHEM 130 Lecture Notes - Lecture 25: Heat Capacity, Calorimetry, Enthalpy

Related Questions