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gayaneshka [121]
3 years ago
11

A 75.0 g sample of dinitrogen monoxide is confined in a 3.q L vessel. What is the pressure( in atm) at 115 celsius

Chemistry
1 answer:
Nonamiya [84]3 years ago
7 0
Data Given:
                  Pressure  =  P  =  ?

                  Volume  =  V  =  3.0 L

                  Temperature  =  T  =  115 °C + 273  =  388 K

                  Mass  =  m  =  75.0 g

                  M.mass  =  M  =  44 g/mol

Solution:
              Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
                                      P V  =  n R T
Solving for P,
                                      P  =  n R T / V      ------ (1)
Calculating Moles,
                                      n  =  m / M

                                      n  =  75.0 g / 44 g.mol⁻¹

                                      n  =  1.704 mol

Putting Values in Eq. 1,

                    P  =  (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

                    P  =  18.08 atm
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CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9 at 1,000 K

Calculate Kc for the reaction 2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

<u>Answer:</u> The value of K_c' for the final reaction is 1.936\times 10^{19}

<u>Explanation:</u>

The given chemical equations follows:

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c

We need to calculate the equilibrium constant for the equation, which is:

2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c'=(K_c)^2

We are given:

K_c=4.4\times 10^9

Putting values in above equation, we get:

K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}

Hence, the value of K_c' for the final reaction is 1.936\times 10^{19}

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