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RSB [31]
2 years ago
15

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.

Chemistry
1 answer:
Eduardwww [97]2 years ago
4 0

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

\large{\boxed{\bold{\Delta E=\Delta H-P.\Delta V}}}

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

brainly.com/question/10600048

an exothermic reaction

brainly.com/question/1831525

 as endothermic or exothermic

brainly.com/question/11419458

an exothermic dissolving process

brainly.com/question/10541336

Keywords: exothermic, endothermic, enthalpy, internal energy, heat

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Write packing efficiency of fcc ,BCC, SCC and formula also . ?​
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❖ <u>Packing Efficiency</u> ❖

➪ The percentage of total space occupied by particles is called <u>packing efficiency</u>.

\\ \qquad{\rule{200pt}{3pt}}

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

❖ Packing efficiency of simple cubic structure (SCC).❖

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 1 atom (sphere). So, the total volume of sphere will be :

:  : \implies \sf \: 1 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi r^3

  • Volume of unit cell = a³

  • Volume of unit cell = (2r)³

  • Volume of unit cell = 8r³

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪ Substituting the known values in the formula, we get the following results:

:  : \implies \rm \: {\dfrac{\dfrac{4}{3}\pi r^3}{8r^3} \times 100} \\ \\ :  : \implies   \dfrac{\pi}{6} \times 100 \\\\\   :  : \implies \bf {52.4 \%}

❒ Hence, the packing efficiency of simple cubic structure is 52.4%.

\\ \qquad{\rule{200pt}{3pt}}

➪ Packing efficiency of cubic close packing (SCC)/face centred cubic structure (FCC).

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

: : \implies \rm 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3

  • Volume of unit cell = a³

Volume of unit cell = (2√2r)³

Volume of unit cell = 16√2r³

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪Substituting the known values in the formula, we get the following results:

:   : \implies \rm {\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 }\\\\  \implies \rm \dfrac{\pi}{3\sqrt{2}} \times 100 \\\\\  \implies\bf52.4 \%

❖ Hence, the packing efficiency of face centred cubic structure is 74%.

\\ \qquad{\rule{200pt}{3pt}}

❖ Packing efficiency of body cubic structure (BCC).

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

: : \implies  \rm \: 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3

  • Volume of unit cell = a³

Volume of unit cell = (4r/√3)³

Volume of unit cell = 64r³/3√3

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪Substituting the known values in the formula, we get the following results:

: : \implies \rm {\dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100} \\\\  \implies \dfrac{3\pi}{8} \times 100 \\\ \\  \bf  \implies \: 68 \%

❒ Hence, the packing efficiency of body centred cubic structure is 68%.

\\ \qquad{\rule{200pt}{3pt}}

6 0
2 years ago
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