For the reaction below, KP = 1.6 × 106 at 400.0 °C. HCHO2( g ) CO( g ) + H2O( g ) A mixture of CO( g ) and H2O(l ) was prepared in a 2.00 L reaction vessel at 25 °C in which the pressure of CO was 0.177 atm. The mixture also contained 0.391 g of H2O. The vessel was sealed and heated to 400 °C. When equilibrium was reached, what was the partial pressure of the HCHO2 at 400 °C ?

Question

For the reaction below, KP = 1.6 × 106 at 400.0 °C. HCHO2( g ) <--> CO( g ) + H2O( g ) A mixture of CO( g ) and H2O(l ) was prepared in a 2.00 L reaction vessel at 25 °C in which the pressure of CO was 0.177 atm. The mixture also contained 0.391 g of H2O. The vessel was sealed and heated to 400 °C. When equilibrium was reached, what was the partial pressure of the HCHO2 at 400 °C ?

DrBob222 · Answer

mol H2O = 0.391/18 = about 0.02 mols but you need to do it more accurately. Then use PV = nRT to calculate P at 400 C. I get about 0.6 atm.
P CO = 0.177 atm at 25. Convert to P at 600. I obtained approximately 0.4 atm.
........HCHO2 ==> CO + H2O
I........0........0.4...0.6
C........x........-x.....-x
E........x......0.4-x..0.6-x
Kp = 1.6E6 = pCO*pH2O/pHCHO2
Substitute the E line and from the ICE chart and solve for x = pHCHO2 in atm. Post your work if you get stuck.

victor · Answer

0.200