Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Part A Which of the following processes is exothermic? CO 2 (g) ? CO 2 (l) H 2 O

ID: 843242 • Letter: P

Question

Part A

Which of the following processes is exothermic?

CO2 (g) ? CO2 (l)

H2O (s) ? H2O (l)

Br2 (l) ? Br2 (g)

I2 (s) ? I2 (g)

C2H5OH (l) ? C2H5OH (g)

Part B

A particular system is doing work on the surroundings as well as gaining heat from the surroundings. What can be said about the values of q (heat) and w (work) for thie system?

q is positive and w is positive

q is negative and w is positive

nothing can be determined about q and w

q is positive and w is negative

q is negative and w is negative

  

Part C

The enthalpy change for the reaction A + 2B ? 3C can be determined using the following equations:

(1)     2 Z + C ? X          ?H = 15 kJ

(2)     2 B + 3 X ? 3 Z    ?H = 30 kJ

(3)     6 X ? A + 9 Z       ?H = 20 kJ

Part D

To do this, which of the following changeshould be made to equation 2?

No changes should be made.

The equation should be flipped.

The equation should be flipped and multiplied by a factor of 3.

The equation should be multiplied by a factor of 3.

The equation should be multiplied by a factor of 1/3.

Part E

Consider:

2 B5H9 (g) + 12 O2 (g) ? 5 B2O3 (g) + 9 H2O (g)      ?H?reaction= - 8687 kJ.

What is the enthalpy change when 0.2500 moles of B5H9 are combusted?

+1737 kJ

-2171 kJ

-1086 kJ

-6248 kJ

+1422 kJ

CO2 (g) ? CO2 (l)

H2O (s) ? H2O (l)

Br2 (l) ? Br2 (g)

I2 (s) ? I2 (g)

C2H5OH (l) ? C2H5OH (g)

Explanation / Answer

Part.A) CO2 (g) ==> CO2 (l)

Part.B) q is positive and w is negative

Part C)

2 Z + C ? X.............1)                           A + 9 Z ==>6X   delH=-20kj.......a)

2 B + 3 X ? 3 Z................2)                    2B+3X==>3Z    delH=+30kj.........b)

6 X ? A + 9 Z....................3)                   3X==>6Z+3C      delH=-45kj.........c)

summing a, b and c gives A + 2B==>3C delH=-35kj

part D) No changes should be made to equation 2)

part E) -1086 kJ (i.e by dividing all the terms of equation by 8)