Question

What is the enthalpy change (in kJ) for burning 30.0 g of methyl alcohol? What is the sign of the change in enthalpy for an exothermic reaction? Hint, when a reaction is exothermic,what is the sign (positive or negative) for qsurr? Remember, energy is conserved. So, ΔH +qsurr = 0. In other words when one is positive the other is negative.) Remember that the 727 kJ is per mole of CH3OH.)

Answer 1

The enthalpy change for the combustion of 30.0 g of methanol is -680.112 kJ, indicating an exothermic reaction where heat is released to the surroundings.

Step-by-step explanation:

The enthalpy change for burning methyl alcohol (methanol, CH3OH) can be found using stoichiometry and the heat of combustion. Since methyl alcohol has a known heat of combustion of 727 kJ/mol, we will need to first determine how many moles of methanol 30.0 g represents.

The molar mass of methanol (CH3OH) is approximately 32.04 g/mol. To convert from grams to moles, we divide the mass of methanol by its molar mass:

30.0 g CH3OH * (1 mol CH3OH / 32.04 g CH3OH) = 0.936 mol CH3OH

Next, since the combustion of 1 mole of methanol releases 727 kJ of energy, the combustion of 0.936 mol would release:

0.936 mol CH3OH * (727 kJ/mol) = 680.112 kJ

Since this is an exothermic reaction, the sign of the enthalpy change (∆H) is negative, indicating that energy is being released to the surroundings. Thus, the enthalpy change for the burning of 30.0 g of methanol is -680.112 kJ. The sign of ∆H corresponds to the opposite of the sign for qsurr, the heat of the surroundings, because of the conservation of energy. Therefore, if ∆H is negative, qsurr would be positive, signifying that the surrounding environment has absorbed the released heat.

Answer 2

The enthalpy change for burning 30.0 g of methyl alcohol is -679.4 kJ. The change in enthalpy for an exothermic reaction is negative (-).

Step-by-step explanation:

The enthalpy change (ΔH) for burning 30.0 g of methyl alcohol can be calculated using the given information.

The enthalpy change for the combustion of methyl alcohol (CH3OH) is given as 727 kJ/mol. To find the enthalpy change for 30.0 g of methyl alcohol, we need to convert grams to moles:

30.0 g CH3OH * (1 mol CH3OH / 32.04 g CH3OH) = 0.936 mol CH3OH

Now, we can calculate the enthalpy change:

ΔH = 0.936 mol CH3OH * (-727 kJ/mol) = -679.4 kJ

Therefore, the enthalpy change for burning 30.0 g of methyl alcohol is -679.4 kJ.

The sign of the change in enthalpy for an exothermic reaction is negative (-). When a reaction is exothermic, the heat released by the system is transferred to the surroundings, resulting in a decrease in the overall enthalpy of the system.