Question 1 (1 point) In each of the three reactions between NaOH and HCl, the si

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Question 1 (1 point)

In each of the three reactions between NaOH and HCl, the sign of q for the water was positive. This means the the sign of q for the reaction was ______ and the reaction was ______.

Question 1 options:

negative, exothermic

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Question 2 (1 point)

If reacting 50 mL of 0.100 M NaOH with 50 mL 0.100 HCl resulted in a Hrxn= - 50 kJ what would be the new H if 100 mL of 0.100 M NaOH were reacted with 100 mL 0.100 HCl?

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Question 3 (1 point)

If reacting 50 mL of 0.100 M NaOH with 50 mL 0.100 HCl resulted in a Hrxn= - 50 kJ what would be the new H if 50 mL of 0.200 M NaOH were reacted with 50 mL 0.200 HCl?

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Question 4 (1 point)

Using the reaction below:

2 CO2(g) + 2 H2O(l) C2H4(g) + 3 O2(g) Hrxn= +1411.1 kJ

What would be the heat of reaction for this reaction?

0.5 C2H4(g) + 1.5 O2(g) CO2(g) + H2O(l) Hrxn= ??? kJ

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Question 5 (1 point)

What would be the Hrxn for C2H4(g) + H2O(l) C2H5OH(l) ?

2 CO2(g) + 2 H2O(l) C2H4(g) + 3 O2(g) H1= +1411.1 kJ

2 C2H5OH(l) + 6 O2(g) 4 CO2(g) + 6 H2O(l) H2= -2735.0 kJ

Also, would the reaction be exothermic or endothermic?

Question 5 options:

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a) positive, endothermic b) negative, endothermic

negative, exothermic

d) positive, exothermic

Explanation / Answer

1)
SINCE WATER absorbes heat, reaction releases it
Answer: c) negative, exothermic

2)
initially,
number of moles of acid/base reacted = 0.1 M * 50 mL = 5 mmol
now,
number of moles of acid/base reacted = 0.1 M * 100 mL = 10 mmol

since number of moles of acid/base reacting have doubles
H will also double

H = -50 KJ * 2 = -100 KJ
Answer : a) - 100 kJ

I am allowed to answer only 1 question at a time

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