1)Calculate the partial pressure (in atm) of Br 2 at equilibrium when 2.35 atm o
ID: 984450 • Letter: 1
Question
1)Calculate the partial pressure (in atm) of Br2 at equilibrium when 2.35 atm of C2H5Br and 2.35 atm of HBr react at 2500 K according to the following chemical equation:
Report your answer to three significant figures in scientific notation.
2)Calculate the partial pressure (in atm) of CO2 at equilibrium when 22.7 g of PbCO3 decomposes at 500 K according to the following chemical equation:
Report your answer to three significant figures in scientific notation.
Explanation / Answer
C2H5Br(g) + HBr (g) C2H6 (g) + Br2 (g)
Kp = pC2H6*pBr2 / pC2H5Br*pHBr
0.0595 = x*x / ((2.35-x)(2.35-x))
x = 0.461 atm
partial pressure (in atm) of Br2 at equilibrium = 0.461 atm
PbCO3(s) PbO(s) + CO2(g)
No of mol of PbCO3 = 22.7/267.21 = 0.085 mol
Kp = pCO2
pCo2 = 0.0269 atm
but for 0.085 mol reaction
pCo2 = 0.0269*0.085 = 0.00229 atm
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