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3 years ago

8

Consider the following equations. 3 A + 6 B → 3 D, ΔH = -446 kJ/mol E + 2 F → A, ΔH = -107.9 kJ/mol C → E + 3 D, ΔH = +61.9 kJ/m

ol Suppose the first equation is reversed and multiplied by 1/6, the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction? chemPadHelp

1 answer:

Lisa [10]3 years ago

6 0

Answer:

F + C/2 → A + B +D, ΔH +400 kJ/mol

Explanation:

3 A + 6 B → 3 D, ΔH = -446 kJ/mol

first equation is reversed and multiplied by 1/6

1/6(3 D → 3 A + 6 B) , ΔH = +(446) kJ/mol

( D/2 → A/2 + B) , ΔH = +(446) kJ/mol

E + 2 F → A, ΔH = -107.9 kJ/mol

second equation is divided by 2,

(E/2 + F → A/2), ΔH = -107.9 kJ/mol

C → E + 3 D, ΔH = +61.9 kJ/mol

third equation is divided by 2,

(C/2 → E/2 + 3/2 D), ΔH = +61.9 kJ/mol

the three adjusted equations are added.

( D/2 → A/2 + B) + (E/2 + F → A/2)+ (C/2 → E/2 + 3/2 D), ΔH = +61.9 kJ/mol + -107.9 kJ/mol +(446) kJ/mol

D/2 +E/2 + F+ C/2 → A/2 + B +A/2 +E/2 + 3/2 D , ΔH +400 kJ/mol

F + C/2 → A + B +D, ΔH +400 kJ/mol

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using the law of the conservation of the linear momentum:

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the linear momentum is calculated by the next equation

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then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

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Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

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$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

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from this last equation we solve for $V_n$ as:

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Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

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