1) Determine the amount of heat in kJ given off when 1.26 x 10^4 g of NO2 are produced according to the equation:
2NO (g) + O2 (g) --> 2NO2 (g)
and delta (H) = 483.6 kJ/mol
2) A sheet of gold weighing 10 g and at a temperature of 18 degrees C is placed on a flat sheet of iron weighing 20 g and at a temperature of 55.6 degrees C. What is the final temperature of the combined metals? (Assume that no heat is lost to the surroundings and the heat gained by the gold must equal the heat lost by the iron.)
The data you gave for #1 shows delta H as +483.6 and that makes it an endothermic reaction so there will be no heat given off. You may want to re-reread/re-type the question.
2. heat lost by Fe + heat gained by Au = 0
(mass Fe x specific heat Fe x (Tfinal-Tinitial) + (mass Au x specific heat Au x (Tfinal-Tinitial) = 0
Substitute the numbers and solve for Tf. That's the onlyh unknown.
elk
1) To determine the amount of heat in kJ given off when 1.26 x 10^4 g of NO2 are produced, we need to use the equation:
2NO (g) + O2 (g) --> 2NO2 (g)
and the value of delta (H) = 483.6 kJ/mol.
First, we need to convert the given mass of NO2 to moles. To do this, we can use the molar mass of NO2, which is 46.01 g/mol.
Number of moles of NO2 = (mass of NO2) / (molar mass of NO2)
= (1.26 x 10^4 g) / (46.01 g/mol)
≈ 273.84 mol
Since the stoichiometric coefficient of NO2 in the balanced chemical equation is 2, the moles of NO2 is halved.
Number of moles of NO2 (actual) = (1/2) x (273.84 mol)
≈ 136.92 mol
Now, we can calculate the total amount of heat released using the equation:
Total heat released = (number of moles of NO2) x (delta H)
= (136.92 mol) x (483.6 kJ/mol)
≈ 66,219.91 kJ
Therefore, approximately 66,219.91 kJ of heat are released when 1.26 x 10^4 g of NO2 are produced.
2) To determine the final temperature of the combined metals, we can use the heat gained by the gold equals the heat lost by the iron.
The heat gained by gold (q₁) can be calculated using the equation:
q₁ = (mass of gold) x (specific heat capacity of gold) x (change in temperature)
The heat lost by the iron (q₂) can be calculated using the equation:
q₂ = (mass of iron) x (specific heat capacity of iron) x (change in temperature)
Since no heat is lost to the surroundings, the heat gained by gold is equal to the heat lost by iron, so:
q₁ = q₂
(mass of gold) x (specific heat capacity of gold) x (change in temperature of gold) = (mass of iron) x (specific heat capacity of iron) x (change in temperature of iron)
(10 g) x (specific heat capacity of gold) x (final temperature - 18°C) = (20 g) x (specific heat capacity of iron) x (final temperature - 55.6°C)
We can rearrange the equation to solve for the final temperature:
10 x (specific heat capacity of gold) x (final temperature - 18) = 20 x (specific heat capacity of iron) x (final temperature - 55.6)
(10 x specific heat capacity of gold - 20 x specific heat capacity of iron) x final temperature = 20 x specific heat capacity of iron x 55.6 - 10 x specific heat capacity of gold x 18
final temperature = [(20 x specific heat capacity of iron x 55.6) - (10 x specific heat capacity of gold x 18)] / (10 x specific heat capacity of gold - 20 x specific heat capacity of iron)
To solve for the final temperature, we need the specific heat capacities of gold and iron. The specific heat capacity of gold is 0.129 J/g°C, and the specific heat capacity of iron is 0.449 J/g°C.
Substituting the known values:
final temperature = [(20 x 0.449 x 55.6) - (10 x 0.129 x 18)] / (10 x 0.129 - 20 x 0.449)
= [4.99 - 2.3238] / (1.29 - 8.98)
= 2.6662 / (-7.69)
≈ -0.3471°C
Therefore, the final temperature of the combined metals is approximately -0.3471°C.
1) To determine the amount of heat in kJ given off when 1.26 x 10^4 g of NO2 are produced, we need to use the equation q = m * ΔH, where q is the heat in kJ, m is the mass in grams, and ΔH is the enthalpy change in kJ/mol.
First, we need to convert the mass of NO2 from grams to moles. To do this, divide the given mass by the molar mass of NO2, which can be calculated by adding up the atomic masses of nitrogen (N) and oxygen (O):
Molar mass of NO2 = 2 * Atomic mass of N + Atomic mass of O = 2(14.01 g/mol) + 1(16.00 g/mol) = 46.01 g/mol
Number of moles of NO2 = Mass of NO2 / Molar mass of NO2
Number of moles of NO2 = 1.26 x 10^4 g / 46.01 g/mol = 274.7 mol
Now, we can calculate the heat using the equation q = m * ΔH. Since the equation is balanced in terms of moles, and the enthalpy change is given per mole, we can multiply the number of moles by the enthalpy change:
Heat = Number of moles of NO2 * ΔH
Heat = 274.7 mol * 483.6 kJ/mol = 132,956.92 kJ
Therefore, the amount of heat given off when 1.26 x 10^4 g of NO2 are produced is approximately 132,956.92 kJ.
2) To determine the final temperature of the combined metals, we can use the principle of heat transfer, which states that the heat gained by one object is equal to the heat lost by another object in a closed system.
The equation for heat transfer is given by q = m * c * ΔT, where q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, we need to calculate the heat lost by the iron and the heat gained by the gold. Since no heat is lost to the surroundings and the heat gained by the gold must equal the heat lost by the iron, we can set up the equation:
Heat lost by iron = Heat gained by gold
(mass of iron) * (specific heat capacity of iron) * (final temperature - initial temperature of iron) = (mass of gold) * (specific heat capacity of gold) * (final temperature - initial temperature of gold)
Given:
Mass of iron = 20 g
Initial temperature of iron = 55.6°C
Mass of gold = 10 g
Initial temperature of gold = 18°C
The specific heat capacity of iron is typically around 0.45 J/g°C, while the specific heat capacity of gold is approximately 0.13 J/g°C. Note that specific heat capacity is typically given in J/g°C.
Now, we can rearrange the equation to solve for the final temperature:
(Mass of iron) * (specific heat capacity of iron) * (final temperature - initial temperature of iron) = (Mass of gold) * (specific heat capacity of gold) * (final temperature - initial temperature of gold)
(20 g) * (0.45 J/g°C) * (final temperature - 55.6°C) = (10 g) * (0.13 J/g°C) * (final temperature - 18°C)
Now, we can simplify and solve for the final temperature. Remember to convert the temperatures to Kelvin by adding 273.15 to each value:
9(final temperature - 55.6) = 2(final temperature - 18)
9final temperature - 501.6 = 2final temperature - 36
7final temperature = 465.6
final temperature = 465.6 / 7 = 66.51°C
Therefore, the final temperature of the combined metals is approximately 66.51°C.