/ 105 Pa/bar 105 Pa/bar The reaction will go to the right. 48.0 Pa 2. (a) Q=105 Pa/bar, = 5.08 x 10-4 < K It was not really necessary to convert Pa to bar, since the units c cancel Н2 Br2 1L 2HBr (b) Initial pressure: 1370 3310 48.0 3310-x 48.0+2x 1370-x Final pressure: Note that 2x Pa of HBr are formed when x Pa of H2 are consumed. (48.0+2x)2 (1370 – x)(3 310 –x) = 7.2 × 10-4 → x=4.50 Pa %3D Pнz = 1366 Pa, РBry 3 3 306 Ра, РнBг PB12 57.0 Pa %3D

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Can you please walk me through how they got 7.2 x 10^ - 4?

/ 105 Pa/bar 105 Pa/bar
The reaction will go to the right.
48.0 Pa
2. (a) Q=105 Pa/bar,
= 5.08 x 10-4 < K
It was not really necessary to convert Pa to bar, since the units c
cancel
Н2
Br2
1L
2HBr
(b)
Initial pressure:
1370
3310
48.0
3310-x
48.0+2x
1370-x
Final pressure:
Note that 2x Pa of HBr are formed when x Pa of H2 are consumed.
(48.0+2x)2
(1370 – x)(3 310 –x)
= 7.2 × 10-4 → x=4.50 Pa
%3D
Pнz
= 1366 Pa,
РBry 3 3 306 Ра, РнBг
PB12
57.0 Pa
%3D
Transcribed Image Text:/ 105 Pa/bar 105 Pa/bar The reaction will go to the right. 48.0 Pa 2. (a) Q=105 Pa/bar, = 5.08 x 10-4 < K It was not really necessary to convert Pa to bar, since the units c cancel Н2 Br2 1L 2HBr (b) Initial pressure: 1370 3310 48.0 3310-x 48.0+2x 1370-x Final pressure: Note that 2x Pa of HBr are formed when x Pa of H2 are consumed. (48.0+2x)2 (1370 – x)(3 310 –x) = 7.2 × 10-4 → x=4.50 Pa %3D Pнz = 1366 Pa, РBry 3 3 306 Ра, РнBг PB12 57.0 Pa %3D
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