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LenKa [72]
3 years ago
15

Consider the reaction h2(g) + cl2(g) ⇀↽ 2 hcl(g), which is exothermic as written. what would be the effect on the equilibrium po

sition of increasing the temperature?
Chemistry
2 answers:
s2008m [1.1K]3 years ago
5 0

Answer:

Backward direction

Explanation:

Exothermic reaction: It is that reaction in which heat is evolved during reaction.

We are given

H_2+Cl_2\rightleftharpoons 2HCl

It is exothermic reaction.

We know that in exothermic reaction   when we increase the temperature then the equilibrium shift in backward direction and when we decrease the temperature then the equilibrium shift in forward direction.

Therefore, when we increasing the temperature then the equilibrium shift in backward direction.

kupik [55]3 years ago
3 0
I think the effect of increasing temperature would be; the equilbrium will shift back wards. Increase in temperature favors backward reaction since the forward reaction is exothermic and the backward reaction is endothermic. Therefore, the equilibrium will shift back wards, and there will be more reactants (H2 and Cl2) compared to the products
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\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL

(b) T_{2} = 95^{o}F

Convert degree Fahrenheit into degree Cesius as follows.

(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL

(c) T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL

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