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May 5, 2016

# Homework Help: chemistry(check my work)

Posted by bme1 on Monday, April 5, 2010 at 10:01pm.

The initial pressure for the compounds involved in the reaction displayed were determined to be P(SO2(g)) = 0.5932 atm, P(O2(g)) = 0.4886 atm, P(SO3(g)) = 0.09287 atm. Calculate the value of the equilibrium constant (Kp) at 1000 K if the equilibrium pressure of O2(g) was 0.4924 atm.

2SO2(g)+O2(g) = 2SO3(g)
0.4924=(0.09287+2x)^2/[(0.5932-x)^2(0.4886)]
• chemistry(check my work) - DrBob222, Monday, April 5, 2010 at 10:21pm

I don't understand what you've done (more exactly, why you have done it). Why the x? at least why the x in the Kp expression. You may want to use x in this manner, for example, as follows:
0.4886 + x = 0.4924 and solve for x as I've explained below.
initial:
SO2 = 0.5932 atm
O2 = 0.4886
SO3 = 0.09287.

change:
If the eq pressure for O2 is 0.4924, that sets the stage for everything else. That means O2 must have changed from 0.4886 to 0.4924 or it must have increased by 0.0038. Now you do the other changes and eq pressures.
SO2 must have increased by 0.0038 (add it to the initial to find equilibrium pressure) and SO3 must have decreased by 2*0.0038. then plug those numbers into Kp = pSO3^2/pSO2*pO2 and solve for Kp.

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