Question

1. A gas is compressed from an initial volume of 5.35 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the compression the gas releases 124 J of heat.

What is the change in internal energy of the gas?

2.  Calculate ΔHrxn for the following reaction:

CaO(s)+CO2(g)→CaCO3(s)

Use the following reactions and given ΔH values:

Ca(s)+CO2(g)+12O2(g)→CaCO3(s), ΔH= -812.8 kJ
2Ca(s)+O2(g)→2CaO(s), ΔH= -1269.8 kJ

Answer 1

1. We know 1 L = 0.001 m^3

Initial volume = 5.35 L = 5.35 * 0.001 m^3 = 0.00535 m^3

Final volume = 1.23 L = 1.23 * 0.001 m^3 = 0.00123 m^3

Pressure = 1.00 atm = 101325 Pa

We know external compression work

W = P∆V = 101325 Pa * (0.00123 - 0.00535) m^3 = - 417.46 Pa m^3 = - 417.46 J

Pa * m^3 = Joule

Also we know ∆U = q - W ----(1)

∆U = change in internal energy, q = heat , W = work

Since heat is releasing so q = -124 J

From(1) ∆U = -124 J - (- 417.46 J) = 293.46 J

∆U = 293.46 J

(2) Given reactions :-

Ca(s) +CO2(g) +1/2 O2(g) ----> CaCO3 (s) ∆H = -812.8 KJ ---(1)

2Ca(s) + O2(g) ----> 2CaO(s) ∆H = -1269.8 KJ ----(2)

Multiply (1) by 2 and add with reverse of (2)

2Ca(s) + 2CO2(g) + O2(g) ---> 2CaCO3(s), ∆H = - 2 * 812.8 KJ

2CaO(s) -----> 2Ca(s) + O2(g) ∆H = + 1269.8 KJ

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2CaO(s) + 2CO2(g) ----> 2CaCO3(s) ∆H = -355.8 KJ

∆Hrxn = -355.8 KJ