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3 years ago

In the important industrial process for producing ammonia (the Haber Process), the overall reaction is:

Chemistry

1 answer:

irina [24]3 years ago

4 0

Answer:

- 50.2 kJ

Explanation:

From the data given, the heat released per 2.0 moles of NH₃ is - 100.4 kJ.

The negative sign means that the reaction is exothermic and the energy is released.

So, the ΔH in kJ of heat released per mole of NH₃ formed is (100.4 kJ/2) = - 50.2 kJ.

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3 years ago

a chemist wishes to mix some pure acid with some water to produce 16L of a solution that is 30% acid how much pure acid and how

Aloiza [94]

Answer: The volume of acid and water that must be mixed will be 4.8 L and 11.2 L

Explanation:

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Volume of mixture = 16 L

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4 0

3 years ago

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

3 years ago

Read 2 more answers