The heat released from the combustion of 0.05 g of white phosphorus increases the temperature if 150g of water from 25 C to 31.5 C. Calculate the enthalpy change for the reaction.

dH = mass H2O x specific heat H2O x (Tfinal-Tinitial)

This will give you the enthalpy change for the reaction as stated in the problem and that is for 0.05g P. Generally these problems don't want that; they want the enthalpy change per mol but this problem doesn't ask for that.

Well, let's get fired up and calculate the enthalpy change for this fiery reaction, shall we?

First, let's calculate the heat released by the combustion of 0.05g of white phosphorus. The specific heat capacity of water is approximately 4.18 J/g°C.

The heat released can be calculated using the formula:
q = m × c × ΔT

Where:
q = heat released
m = mass of water (150g)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (31.5°C - 25°C = 6.5°C)

Let's plug in the values and ignite our calculation:
q = 150g × 4.18 J/g°C × 6.5°C
q ≈ 4,923 J

Now, we can convert the heat released into the enthalpy change (ΔH) by dividing it by the amount of white phosphorus combusted (0.05g):
ΔH = q / m

ΔH = 4,923 J / 0.05g
ΔH ≈ 98,460 J/g

So the enthalpy change for the combustion of white phosphorus is approximately 98,460 J/g. But remember, this value is with respect to the amount of white phosphorus involved in the reaction.

Hope that heated up your interest in thermodynamics!

To calculate the enthalpy change for the reaction, we need to use the equation:

q = mcΔT

where:
q = heat (or enthalpy change)
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature

First, let's calculate the heat released from the combustion of white phosphorus.

Given:
Mass of white phosphorus (mp) = 0.05 g

To find the heat released (q1) from the combustion of white phosphorus, we can use the heat of combustion (ΔHc) for white phosphorus, which is -631.8 kJ/mol:

ΔHc = q1/moles of white phosphorus

To find the moles of white phosphorus, we need to know the molar mass (M) of white phosphorus, which is 123.895 g/mol:

moles of white phosphorus = mp / M

Now we can calculate the heat released from the combustion of white phosphorus (q1):

q1 = ΔHc * moles of white phosphorus

Next, let's calculate the heat absorbed by the water (q2) to increase its temperature:

Given:
Mass of water (mw) = 150 g
Initial temperature of water (Ti) = 25 °C
Final temperature of water (Tf) = 31.5 °C
Specific heat capacity of water (c) = 4.184 J/g·°C

To find the change in temperature (ΔT):

ΔT = Tf - Ti

Now we can calculate the heat absorbed by the water (q2):

q2 = mcΔT

Finally, let's calculate the enthalpy change (ΔH) for the reaction:

ΔH = -q1 / moles of water

To find the moles of water, we need to know the molar mass of water (Mw), which is 18.015 g/mol:

moles of water = mw / Mw

Now we can calculate the enthalpy change for the reaction:

ΔH = -q1 / moles of water

Note: We need to convert kJ to J for the calculations.

I'll calculate each step for you, please provide the necessary values.

To calculate the enthalpy change for the reaction, we will use the formula:

ΔH = q / n

where:
ΔH = enthalpy change (in J)
q = heat released or absorbed (in J)
n = number of moles of the substance undergoing the reaction

First, we need to determine the heat released (q) by the combustion of white phosphorus. We can use the equation:

q = mcΔT

where:
m = mass of water (in g)
c = specific heat capacity of water (approximately 4.184 J/g°C)
ΔT = change in temperature of water (in °C)

Given:
m (mass of water) = 150 g
c (specific heat capacity of water) = 4.184 J/g°C
ΔT (change in temperature of water) = final temperature - initial temperature = 31.5°C - 25°C = 6.5°C

Plugging the given values into the equation, we get:

q = (150 g) * (4.184 J/g°C) * (6.5°C)
q = 4137 J

Now, we need to determine the number of moles of white phosphorus. First, we convert the mass of white phosphorus to moles using its molar mass:

Molar mass of white phosphorus = 31.0 g/mol

Number of moles of white phosphorus = (0.05 g) / (31.0 g/mol) = 0.00161290323 mol

Finally, we can calculate the enthalpy change (ΔH):

ΔH = q / n
ΔH = 4137 J / 0.00161290323 mol
ΔH ≈ 2.564 x 10^6 J/mol

Therefore, the enthalpy change for the combustion of 0.05 g of white phosphorus is approximately 2.564 x 10^6 J/mol.