Question

1. using the following reactions, calculate the heat of formation, δh_f of cs_2 using hess law. show all your work.

i. c(s) + o_2(g) → co_2(g) δh = -393.3 kj
ii. s(s) + o_2(g) → so_2(g) δh = -293.72 kj
iii. cs_2(l) + 3o_2(g) → co_2(g) + 2so_2(g) δh = -1108.76 kj

1. if 0.250 g of fuel increases the temperature of a calorimeter by 20°c, and the calorimeter is calibrated at 7.28 kj/°c, calculate the heat of combustion of fuel per gram. show all your work.

Explanation:

Step1: Calculate heat for problem 1

• First, write the formation reaction of \(CS_2\): \(C(s)+2S(s)\to CS_2(l)\)
• Manipulate the given reactions:
• Multiply reaction (ii) by \(2\): \(2S(s) + 2O_2(g)\to 2SO_2(g)\), \(\Delta H=-2\times293.72=-587.44\ kJ\)
• Reverse reaction (iii): \(CO_2(g)+2SO_2(g)\to CS_2(l)+3O_2(g)\), \(\Delta H = 1108.76\ kJ\)
• Add reaction (i), the multiplied reaction (ii) and the reversed reaction (iii):

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$$ \Delta H_f=(- 393.3)+(-587.44)+1108.76=128.02\ kJ $$

Step2: Calculate heat for problem 2

• Use the formula \(q = C\Delta T\), where \(C = 7.28\ kJ/^{\circ}C\) and \(\Delta T=20^{\circ}C\)

$$q = 7.28\times20 = 145.6\ kJ$$

• Calculate heat of combustion per gram: \(\frac{145.6}{0.250}=582.4\ kJ/g\)

Answer:

1. The heat of formation of \(CS_2\) is \(128.02\ kJ\)
2. The heat of combustion of fuel per gram is \(582.4\ kJ/g\)