2 In a steam-reforming reaction, methane reacts with water to produce hydrogen and carbon monoxide. If 1.56x10-2 mol of methane, 10.4×10-3 mol of water, 2.08x10-2 mol of carbon 4 monoxide, and 7.8x10-3 mol of hydrogen are placed in a 2600 mL steel reactor and heated to 647 °C, will the reaction be at equilibrium or will it proceed to the right to produce CO and H₂ or to the left to form CH4 and H₂O? CH4(g) + H₂O(g) = CO(g) + 3H₂(g) Given Kat 647 °C is 3.12x10-4
2 In a steam-reforming reaction, methane reacts with water to produce hydrogen and carbon monoxide. If 1.56x10-2 mol of methane, 10.4×10-3 mol of water, 2.08x10-2 mol of carbon 4 monoxide, and 7.8x10-3 mol of hydrogen are placed in a 2600 mL steel reactor and heated to 647 °C, will the reaction be at equilibrium or will it proceed to the right to produce CO and H₂ or to the left to form CH4 and H₂O? CH4(g) + H₂O(g) = CO(g) + 3H₂(g) Given Kat 647 °C is 3.12x10-4
General Chemistry - Standalone book (MindTap Course List)
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Chapter14: Chemical Equilibirum
Section: Chapter Questions
Problem 14.31QP: You place 0.600 mol of nitrogen, N2, and 1.800 mol of hydrogen, H2, into a reaction vessel at 450C...
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Transcribed Image Text:2
In a steam-reforming reaction, methane reacts with water to produce hydrogen and carbon
monoxide. If 1.56×10-2 mol of methane, 10.4×10-3 mol of water, 2.08×10-2 mol of carbon
hs
monoxide, and 7.8x10-3 mol of hydrogen are placed in a 2600 mL steel reactor and heated
to 647 °C, will the reaction be at equilibrium or will it proceed to the right to produce CO and
H₂ or to the left to form CH4 and H₂O?
CH4(g) + H₂O(g) = CO(g) + 3H₂(g)
Given Kat 647 °C is 3.12×10-4
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