For the reactions, H2(g)+Cl2(g)----->2HCl(g)+xKJ H2(g)+Cl2(g)----->2HCl(l)+yKJ Which one of the following statement is correct? A)x>y B)x<y C)x=y D)more data required

Answer 1

This is quite a difficult question in that we have to make assumptions: I think the answer is #B#

#H_2(g) + Cl_2(g) rarr 2HCl(g) +Delta_1#
#H_2(g) + Cl_2(g) rarr 2HCl(l) +Delta_2#
The difference between #Delta_1# and #Delta_2# is the process:
#2HCl(l) + Deltararr2HCl(g)#
Energy is used to boil the liquid. #Delta_2# as written should be larger than #Delta_1# (i.e. when the gas is condensed energy is released).
Note that the fact that exothermic reactions (and this one certainly is) are written as NEGATIVE, is a distractor. Here, because the energy is listed as a product (and not as a separate thermoodynamic property), we should be able to say that #Delta_2>Delta_1#. That is we get more energy out when the product is a liquid than when it is a gas.

Others might view this differently (and they might even have a better explanation for the response!).

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Answer 2

The correct statement is B) x < y.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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