A reaction vessel contains 8.06 g of H2 and 64.0 g of O2 at a temperature of 125

Question

A reaction vessel contains 8.06 g of H2 and 64.0 g of O2 at a temperature of 125 C and a pressure of 101 kPa. (a) What is the volume of the vessel? (b) Now, the hydrogen and oxygen are now ignited by a spark, initiating the reaction: 2H2 + O2 2 H2O. This reaction consumes all the hydrogen and oxygen in the vessel. What is the pressure of the resulting water vapor when it returns to its initial temperature of 125 C? (c) What is rms speed of the water molecule in the vessel? (d) What is the internal energy of water vapor in the vessel? (d) How many water vapor molecules are available in the vessel after the spark? (Boltzmann constant, k 1.38 x 1 023 J/K, R-831 /mol. K, Avogadro's number, NA-6022x1023 molecules/mole)

Explanation / Answer

Number of moles required which is given by, n = m / M = [m (H2) + m (O2)] / M (2H2O)

n = [(8.06 g) + (64 g)] / (36 g/mole)

n = (72.06 g) / (36 g/mole)

n = 2 mole

(a) The volume of vessel will be given as -

using an ideal gas law, we have

Pi Vi = n R Ti

where, R = molar gas constant = 8.31 J/mol.K

T = initial temperature = 125 0C = 398.15 K

Pi = initial pressure = 101 x 103 Pa

then, we get

Vi = [(2 mol) (8.31 J/mol.K) (398.15 K)] / (101 x 103 Pa)

Vi = 0.065 m3

(d) How many water vapor molecules are available in vessel after the spark?

N = (6.022 x 1023 molecules/mole) (2 mole)

N = 1.204 x 1024 molecules

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