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vodka [1.7K]
3 years ago
13

How many liters of phosphine are produced when 34 L of hydrogen reacts with an excess of phosphorus under STP?

Chemistry
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:

22.67 L of PH₃

Explanation:

The balanced equation is:

P_4 (s) + 6H_2(g) \to 4PH_3(g)

From the equation:

34 L \times \dfrac{1 \ mol \ of H_2 }{22.4 \ L \ H_2} \times \dfrac{4 \ mol \ of \ PH_3}{6 \ mol \ H_2} \times \dfrac{22.4 \ L \ PH_3}{1 \ mol \ PH_3}

= 22.67 L of PH₃

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Answer:

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Explanation:

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2 years ago
A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

First we have to calculate the equilibrium constant for the reaction.

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\Delta G^o=-RT \ln k

where,

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R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

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Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

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The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

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x=0.168\text{ and }0.716

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The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

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