The following concerns the combustion of nitromethane (CH3NO2 - a popular fuel for top-fuel dragsters) according to the reaction below:. 2 CH3N02 (1) + ž02(g) → 2C02) + 3H20 (1) + N2(9) a. Burning 121 L of nitromethane at constant pressure releases 1.6 x 100 kJ of energy. What is the enthalpy of combustion of this reaction? The density of nitromethane is 1.16 g/cm². b. Assume you transfer heat involved in the previous problem to 1 kL of water that is initially at 25.0 °C. What is the final temperature of the water?

695.8kJ/mol = Enthalpy of combustion

Final temperature = 407.4°C

Explanation:

Taken the reaction:

2CH₃NO₂ + 3/2 O₂ → 2CO₂ + 3H₂O + N₂

The enthalpy of the reaction is the heat released when 1 mole of nitromethane are in combustion.

We need to find the moles presents in 121L of nitromethane using its density and molar mass (61.04g/mol). As these moles releases 1.6x10⁶kJ, we need to find heat released per mole of nitromethane

Mass nitromethane:

121L = 121000cm³ * (1.16g/cm³) = 140360g

Moles:

140360g * (1mol / 61.04g) = 2299.5 moles of nitromethane

Heat released per mole of nitromethane:

1.6x10⁶kJ / 2299.5 mol = 695.8kJ/mol = Enthalpy of combustion

The symbol - is because the heat is released, a + symbol is when the heat is absorbed.

Using:

Q = m×C×ΔT

Where Q is heat (1.6x10⁶kJ = 1.6x10⁹J)

m is mass of water:

1kL = 1000L = 1x10⁶mL = 1x10⁶g

C is specific heat of water: 4.184J/g°C

And ΔT is change in temperature.

Solving for ΔT:

1.6x10⁹J = 1x10⁶g * 4.184J/g°C * ΔT

ΔT = 382.4°C

As initial temperature was 25°C, final temperature is:

382.4°C + 25°C =

407.4°C

c. 75.7

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explanation: