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A. PCl3(g) + Cl2(g) PCl5(g)
B. CO2(g) + H2(g) CO(g) + H2O(g)
C. H2(g) + Br2(g) 2HBr(g)
D. TlCl3(s) TlCl(s) + Cl2(g)
E. 2NOCl(g) 2NO(g)+ Cl2(g)

2007-03-20 04:07:03 · 1 answers · asked by Tracey Lee ♥ 2 in Science & Mathematics Chemistry

1 answers

This works on the principle that, if you drop the pressure,the reaction will go in whatever direction causes it to be increased again. That will be whatever reaction results in more moles of gaseous product than there is gaseous starting material.
D. Has no gaseous starting material but would produce one mole of gaseous product (Cl2). However, since the starting material is all solid, it is not exhibiting much pressure dependence.
E. One mole of gaseous starting material would produce two moles of gaseous product, so that's the one that's going to show the significant pressure dependence I believe.

2007-03-20 04:25:09 · answer #1 · answered by Ian I 4 · 0 0

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