A mole of X reacts at a constant pressure of 43.0 atm via the reaction X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ Before the reaction, the v
olume of the gaseous mixture was 5.00 L. After the reaction, the volume was 2.00 L. Calculate the value of the total energy change, ΔE, in kilojoules.
1 answer:
The value of the total energy change, ΔE, in kilojoules = -61.93 kJ
<h3>Further explanation</h3>Thermochemistry is a branch of chemistry that studies heat changes in chemical reactions.
Thermochemistry is the application of thermodynamics to chemistry.
Thermochemistry is used to predict energy changes in:
- 1. chemical reaction
- 2. change in form
- 3. solution
Energy in (E) is the energy contained in a material
Internal energy is affected by heat changes and work (w)
Delta H reaction (ΔH) is the amount of heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
Enthalpy (H) is a state function which is the sum of the Energy in (E) and work (w).
At constant pressure (P = fixed), the change in entalphy can be formulated:
ΔH = ΔE + P. ΔV
From the problem it can be seen that
ΔH = -75.0 kJ
Pressure, P = 43.0 atm
= 43 atm x 1,01325 .10⁵ Pa / atm
= 4,357.10⁶ Pa
= 4,357.10⁶ N / m²
Volume changes that occur
ΔV = 2.0 L - 5.0 L
= -3.0 L
= -3.0 x 10⁻³ m³
= 1.3071.10⁴ J
= 13,071 kJ
So that:
ΔE = ΔH - PΔV
ΔE = -75.0 kJ - (4,357.10⁶ .- 3.0 x 10⁻³ m3)
ΔE = -75.0 kJ + 13,071 kJ
ΔE = -61.93 kJ
<h3>Learn more</h3>Delta H solution
an exothermic reaction
as endothermic or exothermic
an exothermic dissolving process
Keywords: exothermic, endothermic, enthalpy, internal energy, heat
You might be interested in
Answer:
7.21 × 10⁴ J
Explanation:
Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.
<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>
<em />
We need to calculate the heat required in different stages and then add them.
The moles of ethanol are:
Solid-liquid transition
Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J
where,
ΔHfus: molar heat of fusion
n: moles
Liquid: from -114.5°C to 78.4°C
Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J
where,
c(l): specific heat capacity of the liquid
ΔT: change in the temperature
Liquid-gas transition
Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J
where,
ΔHvap: molar heat of vaporization
Gas: from 78.4°C to 135.3°C
Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J
where
c(g): specific heat capacity of the gas
Total heat required
Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J