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Bess [88]
3 years ago
8

In the presence of excess oxygen, methane gas burns in a constant-pressure system to yield carbon dioxide and water: CH4 (g) 2O2

(g) → CO2 (g) 2H2O(l) ΔH = -890.0 kJ
Calculate the value of q (kJ) in this exothermic reaction when 1.70 g of methane is combusted at constant pressure.
(A) -94.6 kJ
(B) 0.0306 kJ
(C) -0.0106 kJ
(D) 32.7 kJ
(E) -9.46 × 10^4 kJ
Chemistry
1 answer:
Degger [83]3 years ago
7 0

Answer:

The energy released will be -94.56 kJ or -94.6 kJ.

Explanation:

The molar mass of methane is 16g/mol

The given reaction is:

CH_{4}(g) + 2O_{2} (g) --> CO_{2} (g)+ 2H_{2}O(l)

the enthalpy of reaction is given as  ΔH = -890.0 kJ

This means that when one mole of methane undergoes combustion it gives this much of energy.

Now as given that the amount of methane combusted = 1.70g

The energy released will be:

=\frac{energy released by one moleXgiven mass}{molarmass} =\frac{-890X1.7}{16}= -94.56 kJ

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<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

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\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

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<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

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Evaluating the value of 'x', we get:

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Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

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The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

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3 years ago
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Answer:

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We begin from this relation:

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