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Umnica [9.8K]
3 years ago
13

In a coffee-cup calorimeter, 50.0ml of 0.100M AgNO3 and50.0ml of 0.100M HCl are mixed to yield the following reaction:

Chemistry
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

-66.88KJ/mol

Explanation:

It is possible to obtain the heat involved in a reaction using a calorimeter. Formula is:

q = -C×m×ΔT

<em>Where q is heat of reaction, C is specific heat capacity (4.18J/°Cg), m is mass of solution (100.0g) and ΔT is temperature change (23.40°C-22.60°C = 0.80°C)</em>

Replacing:

q = -4.18J/°Cg×100.0g×0.80°C

q = -334.4J

Now, in the reaction:

Ag⁺ + Cl⁻→ AgCl

<em>AgNO₃ as source of Ag⁺ and HCl as source of Cl⁻</em>

Moles that react are:

0.050L× (0.100mol /L) = 0.0050moles

If 0.0050 moles produce -334.4J. Heat of reaction is:

-334.4J / 0.0050moles = -66880J/mol = <em>-66.88KJ/mol</em>

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Answer:

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Explanation:

Data provided as per the question below:-

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CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o
Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

CO_{(g)}    +      2H_2O_{(g)}   ⇄  CH_3OH_{(g)}

The ICE table can be represented as:

                                  CO_{(g)}    +      2H_2_{(g)}   ⇄  CH_3OH_{(g)}

Initial                          0.27             0.49              0.0

Change                      -x                  -2x                 x

Equilibrium               0.27 - x         0.49 -2x          x

We can now say that the concentration of  CH_3OH_{(g)} at equilibrium is x;

Let's not forget that at equilibrium  CH_3OH_{(g)} = 0.11 M

So:

x =  [CH_3OH_{(g)}] = 0.11 M

[CO_{(g)}] = 0.27 - x

[CO_{(g)}] = 0.27 - 0.11

[CO_{(g)}] = 0.16 M

[2H_2_{(g)}] = (0.49 - 2x)

[2H_2_{(g)}] = (0.49 - 2(0.11))

[2H_2_{(g)}] = 0.49 - 0.22

[2H_2_{(g)}] = 0.27 M

K_C = \frac{[CH_3OH]}{[CO][H_2]^2}

K_C = \frac{(0.11)}{(0.16)[(0.27)^2}

K_C = 9.4307

K_C = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0
3 years ago
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