CH3OH(g) ----> CO(g) + 2H2 (g) (delta H =+90.7k)
A) is heat absorbed or released in the course of this reaction?
b) calculate the amount of heat transferred when 45.0g of CH3OH(g)is decomposed by this reaction at constant pressure.
c) for a given sample of CH3OH, the enthalpy change on reaction is25.8kj. how many grams of hydrogen gas are produced? what is thevalue of delta H for the reverse of the previous reaction?
d) how many kilojoules of heat are released when 50.9 g of CO(g)reacts completely with H2O(g) to form CH3OH(g) at constantpressure?
a)DH is +; heat is absorbed.
b)90.7 kJ x (45.0/molar mass CH3OH) = ?
c)(25.8/90.7) x 4 = ?
1)heat is absorbed because delta h is positive
2) first change the grams to moles, then changes the the moles to kj
3)do number two but in reverse
4)
a) The heat is absorbed in this reaction (+90.7 kJ).
b) To calculate the amount of heat transferred, we need to use the equation:
q = m * ΔH
where q is the heat transferred, m is the mass, and ΔH is the enthalpy change.
Substituting the values:
q = 45.0 g * 90.7 kJ/mol = 4081.5 kJ
c) To find the number of grams of hydrogen gas produced, we need to use stoichiometry. From the balanced equation, we can see that 1 mole of CH3OH produces 2 moles of H2. So, the number of moles of H2 can be calculated using the equation:
moles H2 = 25.8 kJ / ΔH = 25.8 kJ / 90.7 kJ/mol = 0.284 mol H2
To convert mol H2 to grams, we use the molar mass of hydrogen gas (H2):
mass H2 = 0.284 mol H2 * 2.02 g/mol = 0.574 g H2
The value of ΔH for the reverse reaction is the negative of the given ΔH. So, for the reverse reaction:
ΔH = -90.7 kJ/mol
d) To calculate the heat released, we need to multiply the moles of CO by the enthalpy change (ΔH), which is -90.7 kJ/mol CO. First, we need to determine the number of moles of CO:
moles CO = mass CO / molar mass CO = 50.9 g / 28.01 g/mol CO = 1.818 mol CO
Heat released = moles CO * ΔH = 1.818 mol CO * -90.7 kJ/mol CO = -164.981 kJ
So, approximately 165 kJ of heat is released when 50.9 g of CO reacts completely with H2O to form CH3OH at constant pressure.
a) To determine if heat is absorbed or released in the course of this reaction, we need to analyze the sign of the enthalpy change (delta H). In this case, the given delta H is +90.7 kJ.
Since the delta H is positive, it means that the reaction is endothermic. This implies that heat is absorbed during the reaction.
b) To calculate the amount of heat transferred when 45.0 g of CH3OH(g) is decomposed, we can use the equation:
q = m * delta H
Where:
q = heat transferred (in joules)
m = mass of substance (in grams)
delta H = enthalpy change (in joules or kilojoules)
First, convert the mass of CH3OH from grams to moles using its molar mass. The molar mass of CH3OH is calculated by adding up the atomic masses of each element (12.01 g/mol + 1.01 g/mol + 16.00 g/mol + 1.01 g/mol = 32.04 g/mol).
Number of moles of CH3OH = mass / molar mass
Number of moles of CH3OH = 45.0 g / 32.04 g/mol ≈ 1.40 mol
Now, we can calculate the heat transferred:
q = m * delta H
q = 1.40 mol * 90.7 kJ/mol
Therefore, the amount of heat transferred when 45.0 g of CH3OH(g) is decomposed is approximately:
q ≈ 1.40 mol * 90.7 kJ/mol = 126.98 kJ (or 1.27 x 10^5 J)
c) To find the number of grams of hydrogen gas produced based on the given enthalpy change, we need to use the stoichiometry of the reaction. According to the balanced equation, the molar ratio of CH3OH to H2 is 1:2. This means that for every 1 mole of CH3OH reacted, 2 moles of H2 are produced.
Let's say the number of moles of H2 produced is x. Using the molar ratio, we can set up a proportion:
1 mol CH3OH / 2 mol H2 = 25.8 kJ / x mol H2
Next, rearrange the equation for x:
x mol H2 = (2 mol H2 * 25.8 kJ) / 1 mol CH3OH
x mol H2 = 51.6 kJ / 1 mol CH3OH
Now, we need to calculate the number of grams of H2 produced. We can use the molar mass of H2 (2.02 g/mol).
Mass of H2 = x mol H2 * molar mass of H2
Mass of H2 = (51.6 kJ / 1 mol CH3OH) * (2.02 g H2 / 1 mol H2)
Therefore, the number of grams of hydrogen gas produced is:
Mass of H2 ≈ 104 g
To find the value of delta H for the reverse of the previous reaction, we can simply change the sign. Since the previous reaction had a positive delta H of +90.7 kJ, the reverse reaction would have a delta H of -90.7 kJ.
d) To calculate the amount of heat released when 50.9 g of CO(g) reacts completely with H2O(g) to form CH3OH(g), we can use the same equation as in part b:
q = m * delta H
First, we need to determine the amount of CH3OH produced. Again, we use the stoichiometry of the equation, which states that 1 mole of CO reacts to produce 1 mole of CH3OH. Since we are given the mass of CO, we can convert it to moles using its molar mass (28.01 g/mol).
Number of moles of CO = mass / molar mass
Number of moles of CO = 50.9 g / 28.01 g/mol ≈ 1.82 mol
The amount of heat released is given by:
q = m * delta H
q = 1.82 mol * (-90.7 kJ/mol)
Therefore, the amount of heat released when 50.9 g of CO(g) reacts completely with H2O(g) to form CH3OH(g) is approximately:
q ≈ 1.82 mol * (-90.7 kJ/mol) = -165.31 kJ (or -1.65 x 10^5 J)