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andreev551 [17]
4 years ago
10

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3 At this temp, 0.3000 mol of H2 and 0.300 mol o

f I2 were placed in a 1L container to react. What concentration of HI is present at equilibrium? H2+I2<-> 2HI
Chemistry
1 answer:
Vlada [557]4 years ago
4 0
Your answer is: Kc = [HI]² / {[H2] [I2]} 
<span>53.3 = (2x)² / {(0.400M - x)(0.400M - x)} </span>
<span>sqrt(53.3) = 2x / (0.400M - x) </span>
<span>2.92 - 7.30x = 2x </span>
<span>x = 0.314M </span>

<span>[HI] = 2x = 0.628M

:) hopefully this helped. (:</span>
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3 years ago
4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K
swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

Pv =nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 4.3 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

4.3 liter = 4.3\times10^{-3}=4.3\times10^{-3} m^{3}

And initial temperature of balloon, T_{1} = 350 K

Let the final volume of balloon is V_{2}

And as given, final temperature of balloon, T_{2} is 250 K

Now, V_{1} = KT_{1}

4.3\times10^{-3}=K\times350\ (equation\ 1 )

V_{2} = KT_{2}

=K\times250\ (equation 2)

Dividing equation 1 and 2,

\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}

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By cross multiplication:

350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\          = \frac{1075\times10^{-3}}{350} \\          =2.87\times10^{-3}m^{3}

Now, convert it into liter with the help of calculation done above,

2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter

Therefore, volume of the gas in the balloon at 250 K will be  2.87 liter.

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The answer to your question is below.

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