Home /
Expert Answers /
Chemical Engineering /
didnt-understand-these-values-corresponding-5-and-6-can-you-calculate-again-by-writing-thanks-acc-pa891

? didnt understand these values corresponding 5 and 6 can you calculate again by writing thanks according to my previous question teacher

(c) At \( 750^{\circ} \mathrm{C}(1023 \mathrm{~K}) \) for the reaction: \( 2 \mathrm{Zn}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{ZnO}(\mathrm{s}), K_{3}=\frac{1}{P_{0_{2}}} \) \[ \begin{aligned} \Delta G_{3}^{o} & =-118256 \mathrm{cals}=R T \ln p_{\mathrm{O}_{2}} \\ \log p_{\mathrm{O}_{2}} & =-25.3675 \end{aligned} \] Similarly for the reaction (4): \( \mathrm{ZnS}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g})=\mathrm{ZnO}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{~g}) \), we can express: \[ \log p_{\mathrm{O}_{2}}=-12.6507+\frac{2}{3} \log p_{\mathrm{SO}_{2}} \] On substituting the value of \( \log p_{\mathrm{O}_{2}}=-25.3675 \), from Eq. 5 in Eq. 6, we get: \[ \begin{aligned} \log p_{\mathrm{SO}_{2}} & =-19.0773, \\ \therefore p_{\mathrm{SO}_{2}} & =8.37 \times 10^{-20} \text { atm Ans. } \end{aligned} \]

I apologize for the misunderstandi

Live Sessions

Online Lab Report Help

Online Project Report Help

Online Assignment Help

Essay Writing Help

CPM Homework Help

Mortgage Calculator

Electrical Engineering

Civil Engineering

Chemical Engineering

Electronics and Communication Engineering

Mathematics

Physics

Chemistry

Software Works/ Computer Science

Other Subjects

100% Correct Solutions

24/7 Availability

One stop destination for all subject

Cost Effective

Solved on Time

Plagiarism Free Solutions

Confidentiality