A biochemist is investigating the conversion of atmospheric nitrogen to usable form by bacteria that inhabit the root systems of certain legumes and needs to know the pressure in kiloPascals exerted by 1.25g of nitrogen gas in a flask of volume 500 mL at 37 0C.
pV=nRT
p = nRT/V
n = m/M
M (N2) = 28 g/mol
R = 8.314 L kPa K−1 mol−1
n (N2) = 1.25/28 = 0.04 mol
p (N2) = (0.04 x 8.314 x (37 + 273))/ 0.5 = 206 kPa
Comments
2FeS2(g)+1/2O2(g)−→Fe2O3+4SO2(g). TakingtheenthalpyofformationofFeS2(g),Fe2O3 andSO2(g) to be -180, -824 and -297 kJmol−1. If the starting material FeS2 is 0.1, what is the value of enthalpy of reaction and also if the C −v of the calorimeter is 10 kJK−1, what will be the change in the temperature
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