Heat Transfer Reaction: CH3OH to CO + H2

[ahappel]

Heat Transfer!

Consider the following reaction:

CH3OH(g) -> CO(g) + 2H2(g) Change in H= +90.7 kJ

(a) Calculate the amount of heat transferred when 45.0 g of CH3OH(g) are decomposed by this reaction at constant pressure. (in kJ)

(b) If the enthalpy change is 14.0 kJ, how many grams of hydrogen gas are produced? (in g)

(c) How many kilojoules of heat are released when 12.0 g of CO(g) reacts completely with H2 (g) to form CH2OH(g) at constant pressure. (in kJ)

PLEASE HELP!

 

[pikkie]

Sorry can't help

according to the rules in physics forum, we can't help those that didnt show their efforts. We are just guiding, not doing all the homework for you. you must at least show your way of calculations before we can help you.

 

[Blueshift5]

(a) To calculate the amount of heat transferred, we can use the formula:

q = m * ΔH

Where q is the heat transferred, m is the mass of CH3OH(g), and ΔH is the enthalpy change.

Substituting the values given in the problem, we get:

q = (45.0 g) * (+90.7 kJ)

q = 4081.5 kJ

Therefore, 4081.5 kJ of heat would be transferred when 45.0 g of CH3OH(g) are decomposed.

(b) To calculate the amount of hydrogen gas produced, we can use the stoichiometric ratio between CH3OH and H2. From the balanced equation, we can see that for every 1 mole of CH3OH, 2 moles of H2 are produced.

First, we need to convert the given enthalpy change to moles of CH3OH:

ΔH = 14.0 kJ / 90.7 kJ/mol = 0.154 mol CH3OH

Using the stoichiometric ratio, we can calculate the number of moles of H2 produced:

0.154 mol CH3OH * (2 mol H2 / 1 mol CH3OH) = 0.308 mol H2

Finally, we can convert the moles of H2 to grams using its molar mass (2.016 g/mol):

0.308 mol H2 * (2.016 g/mol) = 0.621 g H2

Therefore, 0.621 g of hydrogen gas would be produced with an enthalpy change of 14.0 kJ.

(c) To calculate the amount of heat released, we can use the same formula as in part (a):

q = m * ΔH

However, this time we need to use the enthalpy change for the reverse reaction, which is -90.7 kJ. Also, we need to use the mass of CO(g) since it is the limiting reactant.

First, we need to calculate the moles of CO produced from 12.0 g of CO(g):

12.0 g CO * (1 mol CO / 28.01 g) = 0.428 mol CO

Using the stoichiometric ratio, we can calculate the moles of H2 consumed:

0.428 mol CO * (2 mol