Question

1. A mole of X reacts at a constant pressure of 43.0 atm via the reaction

X(g)+4Y(g)→2Z(g),      ΔH∘=−75.0 kJ

Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L. Calculate the value of the total energy change, ΔE, in kilojoules.

2. A 1.2 kg block of iron at 27 ∘C is rapidly heated by a torch such that 15 kJ is transferred to it. What temperature would the block of iron reach assuming the complete transfer of heat and no loss to the surroundings? If the same amount of heat was quickly transferred to a 880 g pellet of copper at 30 ∘C, what temperature would it reach before losing heat to the surroundings? qcs, Fe(s)cs, Cu(s)===mcsΔT0.450 J/g⋅∘C0.385 J/g⋅∘C

3.To determine whether a shiny gold-colored rock is actually gold, a chemistry student decides to measure its heat capacity. She first weighs the rock and finds that it has a mass of 4.7 g. She then finds that, upon absorption of 52.7 J of heat, the rock undergoes a rise in temperature from 25 ∘C to 57 ∘C. Find the specific heat capacity of the substance comprising the rock.

4. A 55.0-g aluminum block initially at 27.5 ∘C absorbs 725 J of heat. What is the final temperature of the aluminum?

Answer 1

H = E+ PV

dH = dE + d(PV)

this is constant pressure method

dH = dE + pdV

dE = dH - PdV

     = -75 - 43.0 *(5-2)

     = 54 KJ/mol

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heat absorbed = heat transfered from the torch

heat absorbed = mass * specific heat * temp change

1.2 *10³ * 0.45 J/g.^oC * (T2-T1) = 15 *10³

(T2-T1) = 27.78

T1 = 27^oC

T2 = 54.78 ^oC

If 15kJ heat is transfered to pellet of Cu then the rise in temperature will be

880 * 0.385 * (T2-T1) = 15000

T2-T1 = 44.27

Final temp (T2) = 74.27 ^oC

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Heat absorbed = 52.7 J

mass * specific heat * temp difference = 52.7

4.7 * S * (57-25) = 52.7

S = 0.35 J/g.^oC

specific heat of the rock = 0.35 J/g.^oC

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55.0 *0.9 * (T2-T1) = 725

T2-T1= 14.64

T2 = 42.15 ^oC