1. If a substance loses 25J of heat and has 10 J of work done on it, what is the

Question

1. If a substance loses 25J of heat and has 10 J of work done on it, what is the net energy change for the substance? (2 points) 2. A sample of gas occupying 10.0 L is allowed to expand against a constant external pressure of 2.00 atm to 15.0L. What is the energy change in terms of work done? (101.3 J- 1 Latm) (2 points) 3. A stove uses the combustion of methane to produce heat, CHa + 202 > CO2 + 2 HsO ???-890. kJ/mol, If 6.50 grams of methane is burned, what is the amount of heat produced? (2 points) 4. Write a balanced thermochemical equation depicting the formation of one mole of the compound from its elements in their standard states. (1 point) la] NO (8) AH-33.9 kJ/mol Ib) SOs (8) AH-395.2 kJ/mol 5. Find the enthalpy change for the reaction: 2 S02 (8)+02(8) 2 SO,(8) (2 points) Given Substance AHr (kJ/mol) SO2 (8) SO3 (g 296.1 395.2

Explanation / Answer

1.

According to firs law of thermodynamics,

Change in internal energy = q + w

deltaE = - 25 + 10 = - 15 J

2.

Workdone = - P * ( V2 - V1)

W = - 2.00 * ( 15.0 - 10.0)

W = - 10.0 L.atm

W = - 10.0 * 101.3 J

W = - 1013 J

SO, 1013 J of work is done by the system.

3.

From the balanced equation

1 mol (OR) 16 .g of methane gives 890. kJ of heat

Then,

6.50 g. of methane gives 6.50 * 890. / 16.0 = 362. kJ of heat is produced.

4.

(a)

1/2 N2 (g) + O2 (g) ----------> NO2 (g) , deltaH0f = 33.9 kJ/mol

(b)

1/8 S8 (s) + 3/2 O2 (g) ----------> SO3 (g) , deltaH0f = - 395.2 kJ/mol

5.

deltaH0r = 2 * deltaH0f(SO3) - 2 * deltaH0f(SO2) - deltaH0f(O2)

deltaH0r = 2 ( - 395.2 ) - 2 ( - 296.1 ) - 0

deltaH0r = - 198.2 kJ

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