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    2. Steam reacts with methane in the gaseous state to form carbon dioxide and hydrogen.
    (a) State the equation for this reaction.
    (b) 54.06 g of steam and 16.04 g of methane are placed in a 1 dm3 container. When equilibrium has been reached 4.04 g of hydrogen are present in the mixture. Determine the value of Kc for this reaction.
    (c) The reaction is endothermic. State the effect on (i) the amount of hydrogen in the equilibrium mixture and (ii) the value of the equilibrium constant if the reaction is carried out at a higher temperature to the one above.
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    3. Hydrogen and iodine react reversibly in the gaseous state to form hydrogen iodide. A particular equilibrium mixture was found to contain 0.45 mol of iodine, 0.55 mol of hydrogen and 2.4 mol of hydrogen iodide. Calculate the value of the equilibrium constant at this temperature.
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    4. When 1.0 mol of ethanol reacts with 0.5 mol of ethanoic acid at 100 oC the mixture contains 0.42 mol of ethyl ethanoate once equilibrium has been reached. Calculate the value of Kc at 100 oC for this esterification reaction.
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    Kc is calculated by the concentration of the products over the concentration of the reactants, but you have to put the concentrations to their stoichiometric numbers (the big numbers showing the ratios reagents react in).

    2) The equation is 2H2O + CH4 <---> CO2 + 4H2, all in the gaseous state (I used <---> as an analogue for the two half arrows for a reversible reaction.)

    4.04 g of H2 is 2.02 moles of H2. 16.04 grams of CH4 is 16.04/16 = 1.0025 moles. 56.04 grams of water = 56.04/18 = 3.113333.... moles. CO2 isn't given, but using the hydrogen value, there should be 2.02/4= 0.505 moles of CO2.
    Because the container is 1dm^3, and gases fill their container, the amount = concentration (because n/1 = n).

    then, Kc=((.505)x(2.02)^4)/(1.0025 x (3.113333....)^2) = 0.8653 to 4sf

    If the forward is endothermic, more heat = faster forward reaction = more hydrogen in the mixture. Hence, that means more products, so a larger numerator for Kc, and so you'll get a larger value for Kc.
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    (Original post by h3rmit)
    Kc is calculated by the concentration of the products over the concentration of the reactants, but you have to put the concentrations to their stoichiometric numbers (the big numbers showing the ratios reagents react in).

    2) The equation is 2H2O + CH4 <---> CO2 + 4H2, all in the gaseous state (I used <---> as an analogue for the two half arrows for a reversible reaction.)

    4.04 g of H2 is 2.02 moles of H2. 16.04 grams of CH4 is 16.04/16 = 1.0025 moles. 56.04 grams of water = 56.04/18 = 3.113333.... moles. CO2 isn't given, but using the hydrogen value, there should be 2.02/4= 0.505 moles of CO2.
    Because the container is 1dm^3, and gases fill their container, the amount = concentration (because n/1 = n).

    then, Kc=((.505)x(2.02)^4)/(1.0025 x (3.113333....)^2) = 0.8653 to 4sf

    If the forward is endothermic, more heat = faster forward reaction = more hydrogen in the mixture. Hence, that means more products, so a larger numerator for Kc, and so you'll get a larger value for Kc.
    Thank you so much!!! you saved my life.. were you able to solve queastion 3 and 4?
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    (Original post by husras12)
    Thank you so much!!! you saved my life.. were you able to solve queastion 3 and 4?
    They're just the same as 2, really. The only thing to remember is the water as an additional product of esterification.
 
 
 
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