The enthalpy change for the following reaction is 67.7 kJ. N2 (g) + 2 O2 (g) --> 2 NO2 (g) If 700. mL of NO2 gas is collected over water at 20.0 °C and 1.05 atm, calculate the amount of heat absorbed to generate this amount of NO2.

Chemistry: Principles and Practice
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Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
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Chapter5: Thermochemistry
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The enthalpy change for the following reaction is 67.7 kJ.
N2 (g) + 2 O2 (g) --> 2 NO2 (g)
If 700. mL of NO2 gas is collected over water at 20.0 °C and 1.05 atm, calculate the amount
of heat absorbed to generate this amount of NO2.
Transcribed Image Text:The enthalpy change for the following reaction is 67.7 kJ. N2 (g) + 2 O2 (g) --> 2 NO2 (g) If 700. mL of NO2 gas is collected over water at 20.0 °C and 1.05 atm, calculate the amount of heat absorbed to generate this amount of NO2.
Expert Solution
Step 1

Given

Volume of gas , V = 700 mL = 0.700 L

Pressure of gas , P = 1.05 atm

Temperature of gas , T = 20oC = 20 +273.15 = 293.15 K

Moles of NO2 gas , n = PVRT = 1.05 x 0.70.0821 x 293.15  = 0.0305 mol

For formation of 2 mole of NO2 gas , heat absorbed = 67.7 kJ

So, for 0.0305 mole of NO2 gas , heat absorbed = 67.7 x 0.03052 = 1.034 kJ

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