Questions i had trouble with thanks?

Question

1. Phosphorus reacts with excess chlorine according to the following equation.

P4(s) + 10Cl2(g) → 4PCl5(g)

When 6.49 g of PCl5(g) is produced, 11.7 kJ of energy is released.

ΔH for this reaction, in kJ mol-1, is?

2. The equation for the self ionisation of water is given below.

2H2O(l) H3O+(aq) + OH-(aq) ΔH = + 57 kJ mol-1

At 90°C, water is?

Answer 1

1. Phosphorus reacts with excess chlorine

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find moles:

6.49 g of PCl5 @ 208.22 g / mol = 0.03314 moles PCl5

@ P4(s) + 10Cl2(g) → 4PCl5(g):

0.03314 molPCl5 @ 1 mol P4 / 4 mol PCl5 = 0.008285 moles P4

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when  11.7 kJ of energy is released, you have two possible answers for ΔH for this reaction, in kJ / mol:

(A) 11.7kj / 0.03314 moles PCl5 = 353kJ released

one answer:dH = - 353kJ  per mole of PCl5 produced

(B) 11.7kJ/ 0.008285 mol P4 = 1412 kJ released

second possible answer: dH = -1410kJ per mole P4 reacting

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2. The equation for the self ionisation of water is given below.

2H2O(l) -->   H3O+(aq) + OH-(aq) ΔH = + 57 kJ mol-1

At 90°C, water is?

because of the information that you are giving, I think you might be asking about the direction this reaction shifts at higher temperatures, due to the fact that this is endothermic (dH = "+")

if so, then:

your answer is: @ elevated temperatures water is more ionized  into H3O+ & OH- than at lower temperatures